Why are the derivatives of these two equations different? I am trying to implicitly differentiate the following: $\frac {x}{x-y}\ =\ y^2-1 $
I originally multiplied the whole equation by $(x - y)$ and the result of implicitly differentiating the resulting equation ($\ x=xy^2-x-y^3+y $ ) was $\frac {dy}{dx}\ = \frac {2-y^2}{2xy-3y^2+1} $ but then afterwards, when I tried again and differentiated the original expression, $\frac {x}{x-y}\ =\ y^2-1 $, (without manipulating it) I ended up with the correct answer of $\frac {-y}{2y(x-y)^2-x} $.
My question is why does simply changing the form of the equation you're trying to implicitly differentiate (like I have done above) change what you get as $\frac {dy}{dx} $ ?

Contrary to what was said in the comments, you lose no information by your manipulation. In the original equation, you knew $x\ne y$. You don't lose anything by multiplying by $x-y$. You have correctly differentiated both expressions, so we have:
$$
\frac{dy}{dx}=\frac{2-y^2}{2xy-3y^2+1}, \text{ and} \\ \frac{dy}{dx}=\frac{-y}{2y(x-y)^2-x}
$$
We would like to reconcile these two expressions. To do this, we use the equation we started with: $\frac{x}{x-y}=y^2-1$. There are many ways to see that the two expressions are equivalent (on the curve defined by $\frac{x}{x-y}=y^2-1$). One way is to note first that:
$$
\begin{split} \frac{dy}{dx}=\frac{-y}{2y(x-y)^2-x}=\frac{dy}{dx}&=\frac{-y}{(x-y)\left(2y(x-y)-\frac{x}{x-y}\right)} \\ &=\frac{\frac{-y}{x-y}}{2y(x-y)-\frac{x}{x-y}} \\ &=\frac{\frac{-y}{x-y}}{2xy-2y^2-\frac{x}{x-y}} \\ &=\frac{\frac{-y}{x-y}}{2xy-2y^2-(y^2-1)}, \text{ since }\frac{x}{x-y}=y^2-1 \\ &= \frac{\frac{-y}{x-y}}{2xy-3y^2+1} \end{split}
$$
As you can see we are almost at the second expression now. The final part has already been done in the comments. I will repeat it here for completeness. Taking the other expression we have:
$$
\begin{split} \frac{2-y^2}{2xy-3y^2+1} &=\frac{2-\frac{2x-y}{x-y}}{2xy-3y^2+1}, \text{ since }y^2=1+\frac{x}{x-y} \\ &=\frac{\frac{2x-2y-2x+y}{x-y}}{2xy-3y^2+1} \\ &=\frac{\frac{-y}{x-y}}{2xy-3y^2+1}\end{split}
$$
which is what we had above. So both expressions are equivalent. In general you can manipulate your equation before you implicitly differentiate. You do need to be careful if you divide by something that may be $0$. Multiplying by $0$ is fine, always.
