Question: I've started with thinking that if X and Y are perfectly correlated, then it is the same as looking at the correlation of Y and -Y (since the X provides no new information) and thus the correlation is -1.

Is this correct?

Answer: 
I'll treat this as [self-study](/questions/tagged/self-study "show questions tagged 'self-study'"), and I'd encourage you to read [its wiki](https://stats.stackexchange.com/tags/self-study/info) and add the tag.

Your argument is already very good. Here are a few pointers. Feel free to write a comment so we can discuss and work towards a good answer.

* I assume you are looking at Pearson's correlation, right? (Does your argument work for other measures of correlation?)
* What does a perfect correlation mean *graphically*?
* What will $X+Y$ and $X-Y$ look like *graphically* if $X$ and $Y$ are perfectly Pearson-correlated?
$X,Y$ perfectly correlate $\implies Cov(X,Y)=\sqrt{Var(X)Var(Y)}$

$Cov(X+Y,X-Y)=Cov(X,X)-Cov(X,Y)+Cov(Y,X)-Cov(Y,Y)=Var(X)-Var(Y)$

$Var(X+Y)Var(X-Y)=[Var(X)+Var(Y)+2Cov(X,Y)][Var(X)+Var(Y)-2Cov(X,Y)]=[Var(X)+Var(Y)]^2-4Cov(X,Y)Cov(X,Y)=\_{\rho\_{X,Y}=1}=[Var(X)+Var(Y)]^2-4Var(X)Var(Y)=[Var(X)-Var(Y)]^2$

Hence, $\rho\_{X+Y,X-Y}=Cov(X+Y,X-Y)/\sqrt{Var(X+Y)Var(X-Y)}=\pm1$.

As @whuber pointed out, when $Var(X)=Var(Y)$, $\rho\_{X+Y,X-Y}$ is undefined.

`@Dilip Sarwate` has provided the answer in the comment to his post. I added some details.

Actually, if $Var(X)=Var(Y)$, given they are perfectly correlated, $Y=\pm X+b$, hence either $X+Y\equiv0$ or $X-Y\equiv0$. Therefore, $\rho=0$.