Answer
$$
3(x^{2}+y^{2})^{2}=25(x^{2}-y^{2}); \quad\quad (2,1)
$$
The equation of the tangent line at the point $(2,1)$ is:
$$
y =\frac{-2}{11}x+\frac{15}{11}
$$
Work Step by Step
$$
3(x^{2}+y^{2})^{2}=25(x^{2}-y^{2}); \quad\quad (2,1)
$$
Now, we can calculate $\frac{dy}{dx}$ by implicit differentiation,
$$
\begin{aligned}
6\left(x^{2}+y^{2}\right) \frac{d}{d x}\left(x^{2}+y^{2}\right) &=25 \frac{d}{d x}\left(x^{2}-y^{2}\right) \\ 6\left(x^{2}+y^{2}\right)\left(2 x+2 y \frac{d y}{d x}\right) &=25\left(2 x-2 y \frac{d y}{d x}\right) \\
12 x^{3}+12 x^{2} y \frac{d y}{d x}+12 x y^{2}+12 y^{3} \frac{d y}{d x} &=50 x-50 y \frac{d y}{d x} \\
12 x^{2} y \frac{d y}{d x}+12 y^{3} \frac{d y}{d x}+50 y \frac{d y}{d x} &=-12 x^{3}-12 x y^{2}+50 x \\
\left(12 x^{2} y+12 y^{3}+50 y\right) \frac{d y}{d x} &=-12 x^{3}-12 x y^{2}+50 x \\
\frac{d y}{d x} &=\frac{-12 x^{3}-12 x y^{2}+50 x}{12 x^{2} y+12 y^{3}+50 y}
\end{aligned}
$$
To find the slope of the tangent line at the point $(2,1)$,
let $x=2$ and $y=1.$ The slope is
$$
\begin{aligned}
\frac{d y}{d x} &=\frac{-12 (2)^{3}-12 (2) (1)^{2}+50 (2)}{12 (2)^{2} (1)+12 (1)^{3}+50 (1)}\\
&=\frac{-20}{110}\\
&=\frac{-2}{11}\\
\end{aligned}
$$
The equation of the tangent line is then found by using the point-slope form of the equation of a line.
$$
\begin{aligned}
(y-y_{1}) &=m(x-x_{1})\\
(y-(1)) &=(\frac{-2}{11} )(x-(2))\\
11(y-1) &=-2(x-2)\\
11y &=-2x+4+11\\
11y &=-2x+15\\
y &=\frac{-2}{11}x+\frac{15}{11}\\
\end{aligned}
$$
Therefore, the equation of the tangent line at the point $(2,1)$ is:
$$
y =\frac{-2}{11}x+\frac{15}{11}
$$
