Answer
The critical points are $(\frac{9}{2},\frac{3}{2})$ and (0,0).
There is a saddle point at (0,0).
There is a relative maximum at $(\frac{9}{2},\frac{3}{2})$
Work Step by Step
$f(x,y)=x^{2}+4y^{3}-6xy-1$
$f_x(x,y)=2x-6y=0 \rightarrow x=3y$
$f_y(x,y)=12y^{2}-6x=0$
Use the substitution method to solve the system of equations
$f_y(x,y)=12y^{2}-6x=0=0 \rightarrow 12y^{2}-6(3y)=0$
$12y^{2}-18y=0$
$y=\frac{3}{2} \rightarrow x=\frac{9}{2}$ or $y=0 \rightarrow x=0$
The critical points are $(\frac{9}{2},\frac{3}{2})$ and (0,0)
To identify any extrema:
$f_{xx}(x,y)=2$
$f_{yy}(x,y)=24y$
$f_{xy}(x,y)=-6$
For (0,0):
$f_{xx}(0,0)=2$
$f_{yy}(0,0)=0$
$f_{xy}(0,0)=-6$
$D=2\times0 - (-6)^{2}=-36\lt0 \rightarrow$ There is a saddle point at (0,0)
For $(\frac{9}{2},\frac{3}{2})$:
$f_{xx}(\frac{9}{2},\frac{3}{2})=2$
$f_{yy}(\frac{9}{2},\frac{3}{2})=36$
$f_{xy}(\frac{9}{2},\frac{3}{2})=-6$
$=2\times 36- (-6)^{2}=36\gt0$ and $f_{xx}(\frac{9}{2},\frac{3}{2})=2\gt0 \rightarrow$ There is a relative maximum at $(\frac{9}{2},\frac{3}{2})$
