Answer
$\langle -3,3,9\rangle$
Work Step by Step
Given $$3 z^{3}+x^{2} y-y^{2} x=1,\ \ \ \ P=( 1,-1,1)$$
Consider $f(x,y,z)= 3 z^{3}+x^{2} y-y^{2} x-1$
Since
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle\\
&=\left\langle 2xy-y^2, x^2-2xy, 9z^2 \right\rangle\\
\nabla f\bigg|_{(1,-1,1)}&= \left\langle -3,3,9 \right\rangle
\end{align*}
Then the normal vector on the surface $3 z^{3}+x^{2} y-y^{2} x=1$ at $P$ is $\langle -3,3,9\rangle$
