Answer
$$\phi (x,y,z) = xy z^2+K
$$
Work Step by Step
Given $$\mathbf{F}=\left\langle y z^{2}, x z^{2}, 2 x y z\right\rangle$$
We need to find $\phi (x,y )$ such that $$ \frac{\partial \phi }{\partial x}=y z^2,\ \ \ \frac{\partial \phi }{\partial y}=x z^{2},\ \ \ \ \frac{\partial \phi }{\partial z}=2x y z$$
By integration we get
\begin{align*}
\phi (x,y,z)&=xy z^2+C_1\\
\phi (x,y,z)&= xy z^2 +C_2 \\
\phi (x,y,z)&=xy z^2 +C_3
\end{align*}
Then we can choose $$\phi (x,y,z) = xy z^2+K
$$
