Answer
We show that $f$ cannot satisfy Clairaut's Theorem, therefore there does not exist any function $f\left( {x,y} \right)$ such that $\frac{{\partial f}}{{\partial x}} = xy$ and $\frac{{\partial f}}{{\partial y}} = {x^2}$.
Work Step by Step
Assume $\frac{{\partial f}}{{\partial x}} = xy$ and $\frac{{\partial f}}{{\partial y}} = {x^2}$.
We take the mixed derivatives and obtain:
$\frac{{{\partial ^2}f}}{{\partial y\partial x}} = x$ ${\ \ \ }$ and ${\ \ \ }$ $\frac{{{\partial ^2}f}}{{\partial x\partial y}} = 2x$
Since $\frac{{{\partial ^2}f}}{{\partial y\partial x}} \ne \frac{{{\partial ^2}f}}{{\partial x\partial y}}$ for $x \ne 0$, $f$ does not satisfy the Clairaut's Theorem.
However, $\frac{{{\partial ^2}f}}{{\partial y\partial x}}$ and $\frac{{{\partial ^2}f}}{{\partial x\partial y}}$ are continuous functions. Thus, a contradiction. Therefore, we conclude that there does not exist any function $f\left( {x,y} \right)$ such that $\frac{{\partial f}}{{\partial x}} = xy$ and $\frac{{\partial f}}{{\partial y}} = {x^2}$.
