The line $x-2y=2$ has slope $\frac12$, so you want to set the derivative of $\frac{x-1}{x+1}$ equal to $\frac12$ and solve for $x$:
$$\frac2{x^2+2x+1}=\frac12\;.$$
This would actually be just a little easier if you hadn't multiplied out the denominator; then you'd have $$\frac2{(x+1)^2}=\frac12\;,\tag{1}$$ so $(x+1)^2=4$, $x+1=\pm 2$, ...
Added: Let's recap what's going on here. You want some lines that are to be parallel to the line $x-2y=2$. That line has slope $1/2$, so your lines have to have slope $1/2$: that's how you know that they're parallel to the line $x-2y=2$. (If they had a different slope, they'd eventually cross it somewhere.) The derivative $\frac2{(x+1)^2}$ gives the slope of the tangent at any point on the curve $y=\frac{x-1}{x+1}$; we want to know where that slope is $1/2$, so we set the derivative equal to $1/2$ in $(1)$ and solve for $x$. You found that $x=-3$ or $x=1$, so there are two points on the curve where the tangent is parallel to the line $x-2y=2$; one is $(-3,2)$, and the other is $(1,0)$. Now you just find the tangent lines to $y=\frac{x-1}{x+1}$ at those two points.
For the first, for instance, you have $y-2=\frac12\big(x-(-3)\big)=\frac12(x+3)=\frac12x+\frac32$, so $y=\frac12x+\frac72$; I'll leave the other to you.
Note that $dx$ is not a derivative, it's a differential. Same with $dy$.
What you mean, presumably, is that you will take $\frac{dx}{dt}$ and $\frac{dy}{dt}$ for parts (b) and (c). Note that
$$\frac{dy}{dx} = \frac{\quad\frac{dy}{dt}\quad}{\frac{dx}{dt}}$$
so you can use this for (a), (b), and (c). For (a), this can be used to get the slope of the tangent, but to find the equation of the tangent you'll have to do a bit more work. For (b), you want $\frac{dy}{dx}$ to be $0$, so you want $\frac{dy}{dt}=0$ and $\frac{dx}{dt}\neq 0$. For (c), you want $\frac{dy}{dt}\neq 0$ and $\frac{dx}{dt}=0$.
So I have one answer but I can't figure out how to get the other answer so I was seeing if someone could help me out here.
Best Answer
The general method is to use the quotient rule to find the slope of the tangent line at any given $x$, and to solve for $x$ when that slope equals the slope of your given line $x-2y=5$. The resulting equation is a quadratic equation that has two roots. For each of those $x$'s you find the corresponding $y$ and use the point-slope form of a line to get the equation of the two desired tangent lines.
You got the one correct answer, so you got one of the $x$'s, but apparently you missed the other $x$. Did you remember to add a $\pm$ when you solved the quadratic equation? The answer you got comes from the $+$: now use a $-$ to find the other $x$.