Answer
True
Work Step by Step
\[
2 z^{3}-3\left(x^{2}+y^{2}\right) z=F(x, y, z)
\]
We will find the first derivatives:
\[
\begin{array}{c}
F_{x}=\frac{\partial F}{\partial x}=-6 z x \\
F_{y}=\frac{\partial F}{\partial y}=-6 z y \\
F_{y}=\frac{\partial F}{\partial x}=-3\left(x^{2}+y^{2}\right)+6 z^{2}
\end{array}
\]
We will find the second derivative:
\[
\begin{array}{l}
F_{x x}=\frac{\partial^{2} F}{\partial x^{2}}=\frac{\partial(-6 x z)}{\partial y}=-6 z \\
F_{y y}=\frac{\partial^{2} F}{\partial y^{2}}=\frac{\partial(-6 y z)}{\partial y}=-6 z \\
F_{z z}=\frac{\partial^{2} F}{\partial x^{2}}=\frac{\partial\left(6 z^{2}-3\left(x^{2}+y^{2}\right)\right.}{\partial y}=12 z
\end{array}
\]
$0=-6z-6z+12z=F_{x x}+F_{y y}+F_{z z}$
And then F satisfies the equation $0=F_{x x}+F_{y y}+F_{z z}$
