Given $f$ weakly modular of weight $k$ and of level $\Gamma$, show that $f^* = \overline{f(-\bar{z})}$ is also modular of weight $k$ but now of level $\alpha^{-1}\Gamma\alpha$ where $\alpha = \left(\begin{matrix}-1&0\\0&1\end{matrix}\right)$.
I think this is pretty simple and I'm just making an algebraic slip but I can't actually spot where and it's quite frustrating!
My solution so far is as follows: Let $\gamma = \left(\begin{matrix}a&b\\c&d\end{matrix}\right)\in\Gamma$, then $f*(\alpha^{-1}\gamma\alpha z) = \overline{f\left(\left(\begin{matrix}-a&b\\c&-d\end{matrix}\right)(-\bar{z})\right)} = (\overline{c(-\bar{z})-d})^k\overline{f(-\bar{z})} = (-cz-d)^k\overline{f(-\bar{z})}$
which is almost what I need but not quite (there should not be a minus in front of the c essentially). Not sure if this is entirely appropriate for here as it may just be an algebraic slip - but could potentially be me misunderstanding the definitions of weakly modular.
There are three problems here. Two computational, one more significant.
The computational error is that in fact, $\alpha = \alpha^{-1}$, so $$\alpha^{-1}\gamma\alpha = \begin{pmatrix} a&-b\\-c&d\end{pmatrix}.$$
The bigger error is with your second equality: since $\alpha^{-1}\gamma\alpha$ need not be an element of $\Gamma$, there's no reason for this equality to hold.
Finally, notice that in general, $$\frac{a(-z)+b}{c(-z)+d}=\gamma (-z) \ne-\gamma(z)$$
So we have,
$$\begin{align} f^*(\alpha^{-1}\gamma\alpha z) &= \overline{f\left(-\begin{pmatrix} a&-b\\-c&d\end{pmatrix}\overline z\right)}\ne\overline{f\left(\begin{pmatrix} a&-b\\-c&d\end{pmatrix}(-\overline z)\right)}\end{align}.$$
This should give you enough to finish off.