Find Stationary Point(s) for function (two variables): $f(x,y)=3y^3-x^3-2y^2+4x-2y$

Question

Find all stationary pointsfor function $$f(x,y)=3y^3-x^3-2y^2+4x-2y.$$

So far this is what I have $$\frac{\partial f}{\partial x}\left(3y^3-x^3-2y^2+4x-2y\right)=-3x^2-4$$ and $$\frac{\partial f}{\partial y}\left(3y^3-x^3-2y^2+4x-2y\right)=9y^2-4y-2$$

What do I do from here? I know it's somthing along the lines of making them equal to $0$.

Answer 1

You wrote

$\frac{\partial f}{\partial x}\left(3y^3-x^3-2y^2+4x-2y\right)=-3x^2-4$

but it should read

$\frac{\partial f}{\partial x}\left(3y^3-x^3-2y^2+4x-2y\right)=-3x^2+4.$

If $x_0$ is such that $-3x_0^2+4=0$ and if $y_0$ is such that $9y_0^2-4y_0-2=0$,

then $(x_0,y_0)$ is a stationary point of $f.$