Quick Taylor expansion q

Let f(x) = log(1 - x^2). Write down the Taylor series Tf of f, about x = 0, from that of log(1- y).

I'm unsure how to start really, is the question wanting a taylor expansion of log(1-y) and then just substitute in x^2? I'm pretty sure that would be wrong but I started working out the Taylor expansion of log(1-y)

Substituting $x^2$ for $y$ is fine, why wouldn't it be? If you are not convinced, $\ln (1-x^2)=\ln (1-x)+\ln (1+x)$

Let f(x) = log(1 - x^2). Write down the Taylor series Tf of f, about x = 0, from that of log(1- y).

I'm unsure how to start really, is the question wanting a taylor expansion of log(1-y) and then just substitute in x^2? I'm pretty sure that would be wrong but I started working out the Taylor expansion of log(1-y);

You can show that $f^{(k)}(y)=-\frac{1}{(1-y)^k}$

so $ln(1-y) =\sum\limits_{k=0}^{\infty}-\frac{1}{(1-y)^kk!}(y-0)^k$ and now I don't know what to do.

Surely you can't just sub in x^2 because $f^{(k)}(x)\not=-\frac{1}{(1-x^2)^k}$