Answer
$$y = \frac{-1}{2}x+2$$
Work Step by Step
Given $$x y+x^{2} y^{2}=6,\ \ \ \ (2,1) $$
Differentiate both sides with respect to $x$
\begin{align*}
\frac{d}{dx}(x y+x^{2} y^{2})&=\frac{d}{dx}(6)\\
x y^{\prime}+y+2 x y^{2}+2 x^{2} y y^{\prime}&=0\\
[x+2x^2y]y'&= -y-2 x y^{2}
\end{align*}
Then at $(2,1)$
\begin{align*}
m&= \frac{ -1-2 (2)}{2+2(4)}\\
&= \frac{-1}{2}
\end{align*}
Then the tangent line is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-1}{x-2}&=\frac{-1}{2}\\
2(y-1)&= -x+2\\
y-1&= \frac{-1}{2}x+1\\
y&= \frac{-1}{2}x+2
\end{align*}
