Answer
$$
2(x^{2}+y^{2})^{2}=25xy^{2}; \quad\quad (2,1)
$$
The equation of the tangent line at the point $(2,1)$ is:
$$
y =\frac{11}{12}x-\frac{5}{6}.
$$
Work Step by Step
$$
2(x^{2}+y^{2})^{2}=25xy^{2}; \quad\quad (2,1)
$$
Now, we can calculate $\frac{dy}{dx}$ by implicit differentiation, use the product rule and the chain rule as follows:
$$
\begin{aligned}
4\left(x^{2}+y^{2}\right) \frac{d}{d x}\left(x^{2}+y^{2}\right) &=25 \frac{d}{d x}\left(x y^{2}\right) \\ 4\left(x^{2}+y^{2}\right)\left(2 x+2 y \frac{d y}{d x}\right) &=25\left(y^{2}+2 x y \frac{d y}{d x}\right) \\
8 x^{3}+8 x^{2} y \frac{d y}{d x}+8 x y^{2}+8 y^{3} \frac{d y}{d x} &=25 y^{2}+50 x y \frac{d y}{d x} \\
8 x^{2} y \frac{d y}{d x}+8 y^{3} \frac{d y}{d x}-50 x y \frac{d y}{d x} &=-8 x^{3}-8 x y^{2}+25 y^{2} \\
\left(8 x^{2} y+8 y^{3}-50 x y\right)\left(\frac{d y}{d x}\right) &=-8 x^{3}-8 x y^{2}+25 y^{2} \\
\frac{d y}{d x} &=\frac{-8 x^{3}-8 x y^{2}+25 y^{2}}{8 x^{2} y+8 y^{3}-50 x y} \end{aligned}
$$
To find the slope of the tangent line at the point $(2,1)$,
let $x=2$ and $y=1.$ The slope is
$$
\begin{aligned}
\frac{d y}{d x} &=\frac{-8 (2)^{3}-8 (2) (1)^{2}+25 (1)^{2}}{8 (2)^{2}(1)+8 (1)^{3}-50 (2) (1)} \\
&=\frac{-55}{-60}\\
&=\frac{11}{12}\\
\end{aligned}
$$
The equation of the tangent line is then found by using the point-slope form of the equation of a line.
$$
\begin{aligned}
(y-y_{1}) &=m(x-x_{1})\\
(y-(1)) &=(\frac{11}{12} )(x-(2))\\
12(y-1) &=11(x-2)\\
12y &=11x-22+12\\
12y &=11x-10\\
y &=\frac{11}{12}x-\frac{10}{12}\\
y &=\frac{11}{12}x-\frac{5}{6}\\
\end{aligned}
$$
Therefore, the equation of the tangent line at the point $(2,1)$ is:
$$
y =\frac{11}{12}x-\frac{5}{6}
$$
