Answer: We have $f(x,y)=x^3+xy^2-12x^2-2y^2+21x+10.$

For stationary points, $p=\dfrac{\partial f}{\partial x}=0, q=\dfrac{\partial f}{\partial y}=0$.

$p=\dfrac{\partial f}{\partial x}=3x^2+y^2-24x+21=0$......................................(1)

$q=\dfrac{\partial f}{\partial y}=2xy-4y=0 \Rightarrow 2y(x-2)=0$............................(2)

From (2), either $y=0$ or $x=2.$

Putting $y=0$ in (1), we get

 $3x^2-24x+21=0 \Rightarrow x^2-8x+7=0$

$x^2-x-7x+7=0 \Rightarrow x(x-1)-7(x-1)=0. $

$\Rightarrow (x-7)(x-1)=0 \Rightarrow x=1,7$

So, two stationary points are $(1,0)$ …