Answer
$$
y^{2}(x^{2}+y^{2})=20x^{2}; \quad\quad (1,2)
$$
The equation of the tangent line at the point $(1,2)$ is:
$$
y =\frac{8}{9}x+\frac{10}{9}
$$
Work Step by Step
$$
y^{2}(x^{2}+y^{2})=20x^{2}; \quad\quad (1,2)
$$
Now, we can calculate $\frac{dy}{dx}$ by implicit differentiation, use the product rule and the chain rule as follows:
$$
\begin{aligned}
2 y\left(x^{2}+y^{2}\right) \frac{d y}{d x}+y^{2}\left(2 x+2 y \frac{d y}{d x}\right) &=40 x \\
2 x^{2} y \frac{d y}{d x}+2 y^{3} \frac{d y}{d x}+2 x y^{2}+2 y^{3} \frac{d y}{d x} &=40 x \\
2 x^{2} y \frac{d y}{d x}+4 y^{3} \frac{d y}{d x} &=-2 x y^{2}+40 x \\
\left(2 x^{2} y+4 y^{3}\right)\left(\frac{d y}{d x}\right) &=-2 x y^{2}+40 x \\
\frac{d y}{d x} &=\frac{-2 x y^{2}+40 x}{2 x^{2} y+4 y^{3}} \end{aligned}
$$
To find the slope of the tangent line at the point $(1,2)$,
let $x=1$ and $y=2.$ The slope is
$$
\begin{aligned}
\frac{d y}{d x} &=\frac{-2 (1) (2)^{2}+40(1)}{2 (1)^{2} (2)+4 (2)^{3}} \\
&=\frac{32}{36}\\
&=\frac{8}{9}\\
\end{aligned}
$$
The equation of the tangent line is then found by using the point-slope form of the equation of a line.
$$
\begin{aligned}
(y-y_{1}) &=m(x-x_{1})\\
(y-(2)) &=(\frac{8}{9} )(x-(1))\\
9(y-2) &=8(x-1)\\
9y &=8x-8+18\\
y &=8x+10\\
y &=\frac{8}{9}x+\frac{10}{9}\\
\end{aligned}
$$
Therefore, the equation of the tangent line at the point $(1,2)$ is:
$$
y =\frac{8}{9}x+\frac{10}{9}
$$
