$f(x,y) = (3-x)(3-y)(x+y-3)$
I'm trying to find and classify the critical points of the former function. I have that
$$\displaystyle\frac{\partial f}{\partial x} = (3-y)(-x^2-xy+2x+3) \\\displaystyle\frac{\partial f}{\partial y} = (3-x)(-y^2-xy+2y+3)$$
Then both partial derivatives are zero for $(0,0)$ because of the first term of each one.
Now, if we add up the second term of both partial derivatives and multiply the result by $-1$ we have
$$x^2+y^2-2x-2y=0 \equiv (x-1)^2+(y-1)^2-2=0$$
To find extrema in the last circumference I'm using Lagrange multipliers.
Let $g(x,y)=(x-1)^2+(y-1)^2-2=0$. Then taking $L(x,y)=f(x,y)-\lambda g(x,y)$ and its partial derivatives within the system
$$\begin{cases} \displaystyle\frac{\partial L}{\partial x} = 0\\ \displaystyle\frac{\partial L}{\partial y}=0 \\ \displaystyle\frac{\partial L}{\partial \lambda} =0\end{cases}$$
I have
$$\begin{cases} x^2+xy^2-5xy-3x^2+6x-3y+9=2x\lambda+2\lambda\\ x^2+xy^2-5xy-3y^2+6x-3x+9=2y\lambda+2\lambda \\ (x-1)^2+(y-1)^2-2=0\end{cases}$$
Here is where I'm stuck. I'm not sure how to solve this system. Since the first and second equation are alike, if I subtract them I get $3(y^2-x^2)+3(y-x)+2\lambda(y-x)=0$, but I don't see how this helps me to get a solution.