Answer
$$ x^{2}+4 x y+y^{2}=13, \quad(2,1)$$
The equation of the tangent line is
$$ 5y+4 x=13.$$
The equation of the normal line is
$$ 4y-5x=-6 .$$
Work Step by Step
$$ x^{2}+4 x y+y^{2}=13, \quad(2,1)$$
Differentiate both sides this function , we have
$$
\begin{aligned} & 2x+4 x y^{\prime} +4y +2y y^{\prime}=0\\
& 2(x+2y) +2 y^{\prime}( 2 x +y )=0\\
& \Leftrightarrow y^{\prime}=\frac{-( x+2y )}{( 2 x +y )}
\end{aligned}
$$
So the slope of the tangent line at $(2,1)$ is
$$
\left.\frac{d y}{d x}\right|_{x=2, y=1}=\frac{-(2+2(1) )}{( 2(2) +1)} =\frac{-4}{5}
$$
so an equation of the tangent line is
$$y-1=\frac{-4}{5}(x-2), \,\,\ \text{or} \,\,\, 5y+4 x=13.$$
The slope of the normal line is $\frac{5}{4},$
so an equation of the normal line is
$$y-1=\frac{5}{4}(x-2), \,\,\,\ \text{ or} \,\,\, 4y-5x=-6 .$$