#### Answer

$$y =3x-2$$

#### Work Step by Step

Given $$\frac{x}{x+1}+\frac{x}{y+1}=1, \quad(1,1)$$
Differentiate both sides
\begin{align*}
\frac{\frac{d}{dx}\left(x\right)\left(x+1\right)-\frac{d}{dx}\left(x+1\right)x}{\left(x+1\right)^2}+ \frac{(1+y)-xy'}{(1+y)^2} &=0 \\
\frac{1}{\left(x+1\right)^2}+\frac{1}{1+y} - \frac{x}{(1+y)^2} y' &=0\\
y'&= \frac{(1+y)^2}{x\left(x+1\right)^2}+\frac{1+y}{x} \\
\end{align*}
Then
$$m = 3 $$
Hence, the tangent line is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-1}{x-1}&=3\\
y-1&=3(x-1)\\
y&=3x-2
\end{align*}