Find the critical points of the function $$f(x,y)=3x^2y+xy^2-3xy$$ The first partial derivatives are $$f_x(x,y)=6xy+y^2-3y\qquad\text{and}\qquad 3x^2+2xy-3x$$ The critical points are the solutions of the following system of equations $$\left\{\begin{aligned} 6xy+\phantom{2}y^2-3y&=0\\ 3x^2+2xy-3x&=0\end{aligned}\right.$$ We divide this problem into subproblems by factorizing the left-hand sides of the equations: $$\left\{\begin{aligned} y(6x+\phantom{2}y-3)&=0\\ x(3x+2y-3)&=0\end{aligned}\right.$$ We can obviously get each $f_x(x,y)$ and $f_y(x,y)$ equal to zero in two ways. Combination of the possibilities leads to four solutions:

• $\left\{\begin{aligned} x&=0\\ y&=0\end{aligned}\right.$
  
  
• $\left\{\begin{aligned} 3x+2y-3&=0\\ y&=0\end{aligned} \right. \quad \implies \quad$ $\left\{\begin{aligned} x&=1\\ y&=0\end{aligned} \right.$
  
  
• $\left\{\begin{aligned} x&=0\\ 6x+y-3&=0\end{aligned} \right. \quad \implies$ $\quad \left\{ \begin{aligned} x&=0\\ y&=3\end{aligned} \right.$
  
  
• $\left\{\begin{aligned} 6x+\phantom{2}y-3&=0\\ 3x+2y-3&=0\end{aligned} \right. \quad \implies$ $\quad\left\{\begin{aligned} -3y+3&=0\\ 3x+2y-3&=0\end{aligned}\right. \quad \implies$ $\quad \left\{ \begin{aligned} x&=\frac{1}{3}\\ y&=1\end{aligned} \right.$

There are four criticial points: $(0,0)$, $(1,0)$, $(0,3)$ and $(\tfrac{1}{3},1)$.