Let, $ f(x,y)=x^3+3xy^2-3x^2-3y^2+4 \\ \; \\ $

The stationary points are given by $\dfrac{\partial u}{\partial x}=0 $ and $\dfrac{\partial u}{\partial y}=0$

$ \\ \; \\ \therefore \dfrac{\partial u}{\partial x} \;=\; 3x^2+3y^2-6x=0 \; \; \; \therefore x^2+y^2-2x=0 \; \; \ldots(i) \\ $

Also, $ \\ \; \\ \therefore \dfrac{\partial u}{\partial y} …