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
$$frac{partial f}{partial x}left(3y^3-x^3-2y^2+4x-2yright)=-3x^2-4$$  and
$$frac{partial f}{partial y}left(3y^3-x^3-2y^2+4x-2yright)=9y^2-4y-2$$
What do I do from here? I know it's somthing along the lines of making them equal to $0$.

Fred · Answer

You wrote
$frac{partial f}{partial x}left(3y^3-x^3-2y^2+4x-2yright)=-3x^2-4$
but it should read
$frac{partial f}{partial x}left(3y^3-x^3-2y^2+4x-2yright)=-3x^2+4.$
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 $f.$

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

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