Determine whether $ f(x,y)$ exists given the partial derivatives Find a function $z=f(x,y)$  whose partial derivatives are as given, or explain why this is impossible.
We have that $ f_x$ = $ 3x^2y^2-2x$, and $f_y$ = $ 2x^3y+6y$. where $ f_z$ denotes the partial derivative of the function $ f$ with respect to some variable $ z$. 
I believe that given the partial derivatives, there is not a function $ z=f(x,y)$  whose partial derivatives are as given.
Pf:  We will integrate both $f_x$ and $ f_y$ . The integral of $ f_x$  and the integral of $ f_y$  are not equal by calculus. QED. 
Am I correct? 
 A: Seeing that $\partial_y f_x = \partial_x f_y$, we see that such $f$ exists.
To go about finding this $f$:
If $f_x(x,y)=3x^2y^2-2x$, then we know that $$f(x,y)=x^3y^2-x^2+c(y)+A$$ where $c(y)$ is constant in $x$, i.e $c$ does not depend on $x$, and $A$ does not depend on either $x,y$ i.e constant.
Now, $f_y(x,y)=2x^3y+6y$ so, $$f(x,y)= x^3y^2+3y^2+d(x)+B$$ where $d$ is constant in $y$, i.e $d$ is independent of $y$, and $B$ does not depend on $x,y$, i.e it is constant.
Now we get $$f(x,y)=x^3y^2-x^2+c(y)+A
=x^3y^2+3y^2+d(x)+B$$
We see that we put $$c(y)=3y^2, d(x)=-x^2, A=B $$ then we are done.
A: You can answer that without any deeper analysis.
From $f_x=3x^2y^2-2x$ it can be seen that $y^2x^3-x^2+C_1+g(y)$ has to be derivatived with respect to $x$ to obtain $f_x=3x^2y^2-2x$.
From $f_y=2x^3y+6y$ it can be seen that $x^3y^2+3y^2+h(x)+C_2$ has to be derivatived with respect to $y$ to obtain $f_y=2x^3y+6y$.
Of course that we want $y^2x^3-x^2+C_1+g(y)=x^3y^2+3y^2+h(x)+C_2$.
From this it follows $(g(y)-3y^2)+(-x^2-h(x))+(C_1-C_2)=0$
This function of two variables can be everywhere zero only if we have $g(y)=3y^2$ and $-x^2=h(x)$ and $C_1=C_2=C$ where $C$ is any real number.
So the functions you need are $x^3y^2+3y^2-x^2+C$ where $C \in \mathbb R$.
