Let, $ u\;=\; x^3+3xy-15x^2-15y^2+72x $

$ \\ $

Diff. partially w.r.t. x,

$ \therefore \dfrac{\partial u}{\partial x} \;=\; 3x^2+3y-30x-0+72 \\ \; \\ $

In order to find stationary values, we find $\dfrac{\partial u}{\partial x} \;$ and $\dfrac{\partial u}{\partial y} \;$ and equate them to zero

$ \dfrac{\partial u}{\partial x} \;=\; 3x^2+3y-30x-0+72 \;=\; 0 \; \; \; \ldots (i) \\ \; \\ $

Diff. partially w.r.t.y,

$ \dfrac{\partial u}{\partial y} \;=\; 0+3x-0-30y+0 \\ \; \\ $

Equating it to zero,

$ \therefore 3x-30y=0 \\ \; \\ x=10y \; \; \; \ldots (ii) $

Substituting equation (ii) in equation (i),

$ \therefore 3(10y)^2+3y-30(10y)-0+72\;=\;0 \\ \; \\ \therefore 3(100y^2)+3y-300y-0+72\;=\;0 \\ \; \\ \therefore 100y^2+y-100y+24\;=\;0 \\ \; \\ \therefore 100y^2-99y+24\;=\;0 \\ \; \\ \; \\ \; \\ \therefore y\;=\; \dfrac{-(-99) \pm \sqrt{ (-99)^2 -4 \times 100 \times 24 }}{ 2 \times 100} \; = \; \dfrac{ 99 \pm \sqrt{ 9801-9600 }}{ 200} \\ \; \\ \; \\ y \;=\; \dfrac{ 99 \pm \sqrt{ 201 }}{ 200} \\ \; \\ \; \\ \; \\ $

When $ y \;=\; \dfrac{ 99 + \sqrt{ 201 }}{ 200} \; x\;=\; 10 \bigg( \dfrac{ 99 + \sqrt{ 201 }}{ 200} \bigg) \; = \; \dfrac{ 99 + \sqrt{ 201 }}{ 20} $

When $ y \;=\; \dfrac{ 99 - \sqrt{ 201 }}{ 200} \; x\;=\; 10 \bigg( \dfrac{ 99 - \sqrt{ 201 }}{ 200} \bigg) \; = \; \dfrac{ 99 - \sqrt{ 201 }}{ 20} $