Answer
$$\phi (x,y) = \frac{1}{2}x^2+ \frac{1}{2}y^2 +C$$
Work Step by Step
Given $$\mathbf{F}=\langle x, y\rangle$$
We need to find $\phi (x,y)$ such that $$ \frac{\partial \phi }{\partial x}=x,\ \ \ \frac{\partial \phi }{\partial y}=y$$
By integration, we get
\begin{align*}
\phi (x,y)&= \frac{1}{2}x^2+C_1\\
\phi (x,y)&= \frac{1}{2}y^2+C_1\\
\end{align*}
Then we can choose $$\phi (x,y) = \frac{1}{2}x^2+ \frac{1}{2}y^2 +C$$
