The partial derivatives of z=f(x,y)=xy^2-3x^2-y^2+2x+2 are \frac{\partial z}{\partial x}=y^2-6x+2 and \frac{\partial z}{\partial y}=2xy-2y=2y(x-1).
Setting these equal to zero gives a system of equations that must be solved to find the critical points: y^2-6x+2=0, 2y(x-1)=0.
The second equation will be true if y=0, which will lead to the first equation becoming -6x+2=0 so that 6x=2 and x=1/3, making one critical point (x,y)=(1/3,0).
The second equation of the system above will also be true if x=1, which will lead to the first equation becoming y^2-4=0 and y^2=4, making y=\pm 2 and leading to two critical points (x,y)=(1,2), (x,y)=(1,-2).
You didn't ask for this, but we can also classify these critical points as follows:
1) Find the second-order partials: \frac{\partial^{2}z}{\partial x^{2}}=-6, \frac{\partial^{2}z}{\partial y^{2}}=2x-2, and \frac{\partial^{2}z}{\partial x \partial y}=\frac{\partial^{2}z}{\partial y \partial x}=2y.
2) Find the discriminant D=\frac{\partial^{2}z}{\partial x^{2}}*\frac{\partial^{2}z}{\partial y^{2}}-(\frac{\partial^{2}z}{\partial x \partial y})^2=12-12x-4y^2
3) Plug the critical points into the discriminant: D(1,2)=12-12-16=-16, D(1,-2)=12-12-16=-16, and D(1/3,0)=12-4-0=8.
4) Since D(1,\pm 2)=-16<0, the critical points at (1,\pm 2) are saddle points.
5) Since D(1/3,0)=8>0 and \frac{\partial^{2}z}{\partial x^{2}}|_{(1/3,0)}=-6<0, the critical point at (1/3,0) is a local maximum.
Here's a contour map of this function in the xy-plane along with its critical points.
![enter image source here]()
