Answer
$$ f(x,y,z)=\frac{1}{2}x^2+\frac{1}{3}y^3+\frac{1}{4}z^4+K $$
Work Step by Step
Given $$\nabla f=\langle x,y^2,z^3\rangle$$
Since $$ \nabla f=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle=\langle x,y^2,z^3\rangle$$
Then
\begin{align*}
\frac{\partial f}{\partial x}&=x\ \ \ \Rightarrow\ \ f=\frac{1}{2}x^2+C_1\\
\frac{\partial f}{\partial y}&=y^2\ \ \ \Rightarrow\ \ f=\frac{1}{3} y^3+C_2\\
\frac{\partial f}{\partial z}&=z^3\ \ \ \Rightarrow\ \ f=\frac{1}{4}z^4+C_3\\
\end{align*}
Hence
$$ f(x,y,z)=\frac{1}{2}x^2+\frac{1}{3}y^3+\frac{1}{4}z^4+K $$
