Answer
$f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=-6$ and $y=4.$
Work Step by Step
First we will find both partial derivatives. The partial derivative with respect to $x$ is:
$$f_x(x,y)=\frac{\partial}{\partial x}(x^2+4xy+y^2-4x+16y+3)=2x+4y-4$$
The partial derivative with respect to $y$ is:
$$f_y(x,y)=\frac{\partial }{\partial y}(x^2+4xy+y^2-4x+16y+3)=4x+2y+16$$
To find all values of $x$ and $y$ such that $f_x(x,y)=0$ and $f_y(x,y)=0$ we will equate both partial derivatives with $0$:
$$f_x(x,y)=2x+4y-4=0,f_y(x,y)=4x+2y+16=0$$
From the first equation we have: $x+2y-2=0\Rightarrow2y=2-x$. Putting this into the second equation we get:
$$4x+2-x+16=0\Rightarrow3x=-18\Rightarrow x=-6,$$
which gives us:
$$2y=2-x=2-(-6)=8\Rightarrow y=4$$
So, $f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously for $x=-6$ and $y=4.$
