Given V(x, y) = e^x(x cos y - y sin y)
\(\frac { \partial V }{ \partial x } \) = e^x (cos y) +(x cos y - y sin y)e^x
= e^x (cos y + x cos y - y sin y)
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } \) = e^x(0 + cos y - 0) + (cos y +x cos y.- y sin y)e^x
= e^x(2 cos y + x cos y - y sin y) ... (1)
\(\frac { \partial V }{ \partial y } \) = e^x(-x sin y- y cos y- sin y)

\(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } \) = e^x (-x cos y - (-y sin y + cos y) - cos y)
= e^x(- x cos y + y sin y - cos y - cos y)
= e^x (- x cos y + y sin y - 2 cos y) ... (2)
(1)+(2)➝ 
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } \) + \(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } \) = e^x(2 cos y + x cos y - y sin y - x cos y + y sin y - 2 cos y]
= e^x (0) = 0
Hence proved

Given f(x, y) = \(\frac { 3x }{ y+sinx \ } \)
\({ f }_{ x }=\frac { \partial f }{ \partial x } \)
= \(\frac { (y+sin \ x)(3)-3x(cos \ x) }{ { (y+sin \ x) }^{ 2 } } \)
\(=\frac { 3y+3sin \ x-3x \ cos \ x }{ (y+sin{ x })^{ 2 } } \)
\({ f }_{ yx }=\frac { \partial ^{ 2 }f }{ \partial y\partial x } \)
\(=\frac { (y+{ sinx) }^{ 2 }[3]-(3y+3sinx-3xcosx)(2)(y+sinx)(1) }{ { (y+sinx) }^{ 3 } } \)
= \(\frac { (y+{ sinx) }-[3y+3sinx-6y-6sinx+6xcosx] }{ { (y+sinx) }^{ 4 } } \)
\({ f }_{ yx }=\frac { -3y-3sinx+6xcosx }{ { (y+sinx) }^{ 3 } } \) ......(1)
\({ f }_{ y }={ [3x[-1][y+sinx] }^{ -2 }\)
\(=\frac { -3x }{ (y+sin{ x) }^{ 2 } } \)
\(\therefore { f }_{ xy }=-3\left[ \frac { ({ y+sinx) }^{ 2 }(1)-x(2)(y+sinx)(cosx) }{ { (y+sinx) }^{ 4 } } \right] \)
\({ f }_{ xy }=\frac { -3(y+sinx)[y+sinx-2xcosx] }{ { (y+sinx) }^{ 4 } } \)
\({ f }_{ xy }=\frac { -3(y+sinx-2xcosx) }{ ({ y+sinx })^{ 3 } } \) ...(2)
From (1) and (3)
f_xy = f^yx

First we calculate  \(\frac { { \partial }w }{ { \partial x } } (x,y)=\frac { { \partial }(xy) }{ { \partial x } } +\frac { \partial \left( \frac { { e }^{ y } }{ { y }^{ 2 }+1 } \right) }{ \partial x } \)
This gives  \(\frac { { \partial }^{ }w }{ { \partial x } } \) (x, y) = y + 0 and hence \(\frac { { \partial }^{ 2 }w }{ \partial y\partial x } \) (x, y) = 1 On the other hand,
\(\frac { { \partial }w }{ { \partial y } } (x,y)=\frac { { \partial }(xy) }{ { \partial y } } +\frac { \partial \left( \frac { { e }^{ y } }{ { y }^{ 2 }+1 } \right) }{ \partial y } \)
\(=x+\frac { \left( { y }^{ 2 }+1 \right) { e }^{ y }-{ e }^{ y }2y }{ \left( { y }^{ 2 }+1 \right) } \)
Hence, \(\frac { { \partial }^{ 2 }w }{ { \partial x\partial y } } \) (x, y) = 1

First we shall calculate \(\frac { \partial f }{ \partial x } \) (x, y). Note that f is a sum of two functions and so
\(\frac { \partial f }{ \partial x } =\frac { \partial }{ \partial x } sin({ xy }^{ 2 })+\frac { \partial }{ \partial x } \left( { e }^{ { x }^{ 3 }+5y } \right) \)
\(=cos(x{ y }^{ 2 })\frac { \partial }{ \partial x } (x{ y }^{ 2 })+{ e }^{ { x }^{ 3 }+5y }\frac { \partial }{ \partial x } \left( { x }^{ 3 }+5y \right) \)
= cos(xy² ) y² + \({ e }^{ { x }^{ 3 }+5y }\) 3x²
Similarly,
\(\frac { { \partial }^{ }f }{ { \partial y\ } } =\frac { \partial }{ \partial y } sin(x{ y }^{ 2 })+\frac { \partial }{ \partial y } \left( { e }^{ { x }^{ 3 }+5y } \right) \)
\(=cos(x{ y }^{ 2 })\frac { \partial }{ \partial y } (x{ y }^{ 2 })+{ e }^{ { x }^{ 3 }+5y }\frac { \partial }{ \partial y } \left( { x }^{ 3 }+5y \right) \)
= cos(xy²)2xy + 5\({ e }^{ { x }^{ 3 }+5y }\)
Next we consider,
\(\frac { { \partial }^{ 2 }f }{ { \partial y\partial x } } =\frac { \partial }{ \partial y } \left( \frac { \partial f }{ \partial x } \right) =\frac { \partial }{ \partial y } ({ y }^{ 2 }cos(x{ y }^{ 2 })+3{ x }^{ 2 }{ e }^{ { x }^{ 3 }+5y })\)
\(=\frac { \partial }{ \partial y } ({ y }^{ 2 }cos(x{ y }^{ 2 }))+\frac { \partial }{ \partial y } ({ 3x }^{ 2 }{ e }^{ { x }^{ 3 }+5y })\)
= 2y cos(xy²)+y²(-sin(xy²)2xy) + 3x² \({ e }^{ { x }^{ 3 }+5y }\) 5
= 2y cos(xy²)+2xy³ sin(xy²)+15 x² \({ e }^{ { x }^{ 3 }+5y }\)
Finally,
\(\frac { { \partial }^{ 2 }f }{ { \partial x\partial y } } =\frac { \partial }{ \partial y } \left( \frac { \partial f }{ \partial x } \right) =\frac { \partial }{ \partial y } (cos(x{ y }^{ 2 })2xy+5{ e }^{ { x }^{ 3 }+5y })\)
= -sin(xy²)y²2xy+cos(xy²)2y+5\({ e }^{ { x }^{ 3 }+5y }\) 3x²
= 2y cos(xy²)- 2xy³ sin(xy²)+15x²\({ e }^{ { x }^{ 3 }+5y }\)
Note that we have first used sum rule, then in the next step we have used chain rule. In the third step, product rule is used. Also, we see that f_xy = f_yx Is it a coincidence? or is it always true? Actually, there are functions for which f_xy ≠ f_yz at some points. The following theorem gives conditions under which f_xy = f_yz.

Given (x, y, z) = log (x³ + y³ + z³)
\(\frac { \partial U }{ \partial x } =\frac { 1 }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } { (3x }^{ 2 });\)
\(\frac { \partial U }{ \partial y } =\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \) and
\(\frac { \partial U }{ \partial z } =\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(\therefore \frac { \partial U }{ \partial x } +\frac { \partial U }{ \partial y } +\frac { \partial U }{ \partial z } =\frac { { 3x }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(=\frac { { 3({ x }^{ 2 }+y }^{ 2 }+{ z }^{ 2 }) }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)