`f(x,y)=x^3+x^2y^3-2y^2`
Find Partial derivative, Derivative order = `x^2,y`

Solution:
Now,
`del/(delx)(x^3+x^2y^3-2y^2)`

`=del/(delx)(x^3)+del/(delx)(x^2y^3)-del/(delx)(2y^2)`

`del/(delx)(x^2y^3)=2xy^3+0`

`del/(delx)(x^2y^3)`

`=(del/(delx)(x^2))y^3+x^2(del/(delx)(y^3))`

`del/(delx)(y^3)=0`

`del/(delx)(y^3)`

`=3y^2*del/(delx)(y)`

`=3y^2*0`

`=0`

`=(2x)y^3+x^2*0`

`=2xy^3+0`

`del/(delx)(2y^2)=0`

`del/(delx)(2y^2)`

`=4y*del/(delx)(y)`

`=4y*0`

`=0`

`=3x^2+(2xy^3+0)-0`

`=3x^2+2xy^3+0-0`

`=3x^2+2xy^3`

`:.(del)/(del x)(x^3+x^2y^3-2y^2)=3x^2+2xy^3`

Now, `(del^2)/(del x^2)(x^3+x^2y^3-2y^2)`

`del/(delx)(3x^2+2xy^3)`

`=del/(delx)(3x^2)+del/(delx)(2xy^3)`

`del/(delx)(2xy^3)=2y^3+0`

`del/(delx)(2xy^3)`

`=2*(del/(delx)(x))y^3+2x(del/(delx)(y^3))`

`del/(delx)(y^3)=0`

`del/(delx)(y^3)`

`=3y^2*del/(delx)(y)`

`=3y^2*0`

`=0`

`=2*1y^3+2x*0`

`=2y^3+0`

`=6x+(2y^3+0)`

`=6x+2y^3+0`

`=6x+2y^3`

`:.(del^2)/(del x^2)(x^3+x^2y^3-2y^2)=6x+2y^3`

Now, `(del^3)/(del x^2del y)(x^3+x^2y^3-2y^2)`

`del/(dely)(6x+2y^3)`

`=del/(dely)(6x)+del/(dely)(2y^3)`

`=0+6y^2`

`=6y^2`

`:.(del^3)/(del x^2del y)(x^3+x^2y^3-2y^2)=6y^2`

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