Answer
The critical points are $(126,2646)$ and (0,0).
There is a saddle point at (0,0).
There is a relative maximum at $(126,2646)$
Work Step by Step
$f(x,y)=7x^{3}+3y^{2}-126xy-63$
$f_x(x,y)=21x^{2}-126y=0$
$f_y(x,y)=6y-126x=0 \rightarrow y=21x$
Use the substitution method to solve the system of equations
$f_x(x,y)=21x^{2}-126y=0 \rightarrow 21x^{2}-126(21x)=0$
$21x^{2}-2646x=0$
$x=126 \rightarrow y=2646$ or $x=0 \rightarrow y=0$
The critical points are $(126,2646)$ and (0,0)
To identify any extrema:
$f_{xx}(x,y)=42x$
$f_{yy}(x,y)=6$
$f_{xy}(x,y)=-126$
For (0,0):
$f_{xx}(0,0)=0$
$f_{yy}(0,0)=6$
$f_{xy}(0,0)=-126$
$D=6\times0 - (-126)^{2}=-15876\lt0 \rightarrow$ There is a saddle point at (0,0)
For $(126,2646)$:
$f_{xx}(126,2646)=5292$
$f_{yy}(126,2646)=6$
$f_{xy}(126,2646)=-126$
$D=5292\times 6- (-126)^{2}=158766\gt0$ and $f_{xx}(126,2646)=5292\gt0 \rightarrow$ There is a relative maximum at $(126,2646)$
