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: 
Hint: *In general*, \begin{align}
\rho\_{A,B} &= \frac{\operatorname{cov}(A,B)}{\sqrt{\operatorname{var}(A)\operatorname{var}(B)}},\\
\operatorname{var}(X\pm Y)&= \operatorname{var}(X)+\operatorname{var}(Y)
\pm 2\operatorname{cov}(X,Y),\\
\text{and}\qquad\operatorname{cov}(X+Y,X-Y)&=\operatorname{var}(X)-\operatorname{var}(Y)\end{align} So, work out what $\rho\_{X+Y,X-Y}$ is in general, and in the special case when $Y = aX+b$. You might be surprised at the result.
$X,Y$ 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, $\rho=0$.