937. Find the points at which the function $z=e^{x}\left(x-y^{3}+3 y\right)$ is stationary (i.e., points where the derivative in any direction is zero).
Solution. For the derivative of a function in any direction to be zero at some point $P$, it is necessary and sufficient that all the first-order partial derivatives of the function simultaneously become zero at this point. [According to formula $(a) .1$]

Therefore, by finding the partial derivatives: $z_{x}^{\prime}=e^{x}\left(x-y^{3}+3 y+1\right)$, $z_{u \prime}^{\prime}=3 e^{x}\left(1-y^{2}\right)$ and solving the system of equations $z_{x}^{\prime}=0, z_{t}^{\prime}=0$, we obtain two points: $(-3 ; 1)$ and $(1 ;-1)$, where the function is stationary.