How to find these stationary points in multivariable calculus? $$f(x,y) = 3xy-x^3-y^2$$
So $$f_x = 3y - 3x^2 = 0$$ and $$f_y=3x-2=0$$
I'm confused on how to get the answers: $(0,0)$ and $(3/2, 9/4)$. Can anyone help?
 A: Yeah, as Alberto said, it should be $f_y=3x-2y=0$ so that $y=3x\over 2$ ; and then substituting in the remaining partial derivative $3y-3x^2={9x\over 2} -3x^2= x({9\over 2}-3x)=0$ So we have $x=0$ or $x={3\over 2}$. Then substitute into $y% and you get the required stationary points. 
A: 
Stationary points of the function $~z = f( x,y)~$ are located where the partial derivatives $~\dfrac{\partial f}{\partial x}~$and$~\dfrac{\partial f}{\partial y}~$
  are zero simultaneously. 
i.e., For a stationary point of $~z = f( x,y)~$, $$\dfrac{\partial f}{\partial x}=0\qquad\text{and also}\qquad\dfrac{\partial f}{\partial y}=0~.$$

Here given that $~f(x,y) = 3xy-x^3-y^2~$.
Since $~f(x,y)~$ is polynomial function of$~x~$and$~y~$, so partial derivatives of $~f~$with respect to $~x~$and$~y~$exist.
So according to the above definition $$\dfrac{\partial f}{\partial x}=0\qquad\text{and also}\qquad\dfrac{\partial f}{\partial y}=0$$
$$\implies 3y-3x^2=0\qquad\text{and also}\qquad 3x-2y=0$$
$$\implies \dfrac{9}{2}x-3x^2=0\implies x\left(\dfrac{9}{2}-3x\right)=0\implies x=0,~~\dfrac{3}{2}$$
For $~x=0,~$we have $~y=0~$ and for $~x=\dfrac{3}{2}~$, we have $~y=\dfrac{3}{2}x=\dfrac{9}{4}~$.
So the stationary points of given $~f(x,y)~$ are $~(0,0)~$and$~\left(\dfrac{3}{2},\dfrac{9}{4}\right)~$.
