Finding critical points of multivariable function 
Find the critical points of $f(x,y)=x^y+4xy-y^2-8x-6y$

I found the derivative of the function and got $$f^\prime_x=yx^{y-1}+4y-8  \\ f^\prime_y=\ln x\, x^y+4x-2y-6 $$. I want to find point $(x_0,y_0)$ s.t $f^\prime_x(x_0,y_0)=f^\prime_y(x_0,y_0)=0$. I isolated $x^y$ in both equations and got $x^y=\dfrac{2y+6-4x}{\ln x}=\dfrac{8x-4xy}{y}$, but I can't proceed any further (I get implicit function).  How can I find the critical points?
 A: This function is a pure nightmare ! As said by hardmath, I wonder about a typo such as x^2 instead of x^y.  
Eliminating x from the derivative with respect to x, I found one solution which corresponds to y=1.49336 to which corresponds x=1.85682. This can be obtained using the last equation given by hardmath applying Newton.
A: The system $f_x(x,y)=0$, $f_y(x,y)=0$ cannot be solved explicitly. Maybe there are no solutions at all. With the help of Mathematica one can draw contour plots of $f_x$ and $f_y$ in order to obtain hints where there could be a common zero.
The following figure shows an overlay of the contour plots of $f_x$ and $f_y$ in the rectangle $[1,3]\times[1,2]$. The $0$-level-lines are shown in red. Near $(1.85,1.49)$ there seems to be a common zero $(x_0,y_0)$.

With some trial and error I established
$$f_x(1.858,1.494)=0.00489,\quad f_y(1.858,1.494)=0.00713\ .$$
If you need more accuracy  there are established methods (e.g., Newton's method) for approximating $(x_0,y_0)$ to any desired level of precision.
A: The first equation can be rearranged to give:
$$ x^{y-1} = 8y^{-1} - 4 $$
$$ x = (8y^{-1} - 4)^{\frac{1}{y-1}} $$
$$ \ln x = \frac{\ln(8y^{-1} - 4)}{y-1} $$
and these can in turn be used to eliminate $x$ in the second equation:
$$ (\ln x)x^y + 4x = 2y + 6 $$
$$ [(\ln x) x^{y-1} + 4]x = 2y + 6 $$
$$ \left[ \frac{(8y^{-1} - 4)\ln(8y^{-1} - 4)}{y-1} + 4 \right] 
(8y^{-1} - 4)^{\frac{1}{y-1}} = 2y + 6 $$
So an equation in one variable only is possible, though obviously this is too complicated to offer much hope of a symbolic solution.  It can be simplified somewhat by multiplying through by $(y-1)$ and raising both sides to the $y-1$ power, but it doesn't seem to have a clean answer.
