Let $f\in M_k$ be a modular form of weight $k$ . For $M=\begin{pmatrix} a&b \\c&d\end{pmatrix} \in\Gamma $ (modular group) it holds $$f'(M\tau)=(c\tau+d)^{k+2}f'(\tau)+kc(c\tau+d)^{k+1}f(\tau). $$
Now I want to consider $R_kf(M\tau)$ .
I have that \begin{align} R_kf(M\tau)&=2i((c\tau+d)^{k+2}f'(\tau)+kc(c\tau+d)^{k+1}f(\tau))+ky^{-1}(c\tau+d)^kf(\tau)\\ &=(c\tau+d)^{k+2}\left(2if'(\tau)+\frac{2ikcf(\tau)}{c\tau+d}+\frac{ky^{-1}f(\tau)}{(c\tau+d)^2}\right). \end{align}
My problem is to show that $$\frac{2ikcf(\tau)}{c\tau+d}+\frac{ky^{-1}f(\tau)}{(c\tau+d)^2}=ky^{-1}f(\tau).$$
I will use $z=x+iy$ instead of $\tau$, which is harder to type. To have a clear separation between the different $y$-values, i will write $y(z)$ or $y_z$ for $y=y_z=y(z):=\frac 1{2i}(z+\bar z)$. (This is the reason of the problems above, so i need a notation that distinguishes the different occurrences for $y(z)$ and $y(Mz)$.) So let us restart the computations. $$ \begin{aligned} f(Mz)&=(cz+d)^k\;f'(z)\ , \text{ from $f\in\mathcal M_k(\Gamma)$, so after $\partial/\partial z$} \\[3mm] (Mz)'_z &= \left(\frac{az+b}{cz+d}\right)'_z=\frac{a(cz+d)-c(az+b)}{(cz+d)^2}= \frac{\det M}{(cz+d)^2} =\frac 1{(cz+d)^2} \ , \\ f'(Mz)\cdot (Mz)'_z &=(cz+d)^{k}\;f'(z)+kc(cz+d)^{k-1}\;f(z)\ , \\ f'(Mz)&=(cz+d)^{k+2}\;f'(z)+kc(cz+d)^{k+1}\;f(z)\ . \\[3mm] \text{This implies:}& \\ (R_kf)(Mz) &=2i\;f'(Mz)+ky_{\color{red}{Mz}}^{-1}\;f(Mz) \\ &=2i\;(cz+d)^{k+2}\;f'(z)+2i\;kc(cz+d)^{k+1}\;f(z) +ky_{\color{red}{Mz}}^{-1}\;(cz+d)^k\;f(z) \\ &= (cz+d)^{k+2} \Big[\ 2i\;f'(z)+\frac{2i\;kc}{cz+d}\;f(z) +\frac{ky_{\color{red}{Mz}}^{-1}}{(cz+d)^2}\;f(z) \ \Big] \\ &= (cz+d)^{k+2} \Big[\ \underbrace{2i\;f'(z) +\color{blue}{ky_z^{-1}\; f(z)} }_{(R_kf)(z)}\\ &\qquad\qquad\qquad\qquad -\color{blue}{ky_z^{-1}\; f(z)} +\frac{2i\;kc}{cz+d}\;f(z) +\frac{ky_{\color{red}{Mz}}^{-1}}{(cz+d)^2}\;f(z) \ \Big] \\[3mm] \text{So we need:}& \\ 0 &\overset{(!)}= -\color{blue}{ky_z^{-1}\; f(z)} +\frac{2i\;kc}{cz+d}\;f(z) +\frac{ky_{\color{red}{Mz}}^{-1}}{(cz+d)^2}\;f(z)\ ,\text{ i.e.} \\ 0 &\overset{(!)}= -\color{blue}{y_z^{-1}}(cz+d)^2 +2i\;c\; (cz+d) +y_{\color{red}{Mz}}^{-1}\ . \\[3mm] \text{So we compute:}& \\ y_{\color{red}{Mz}} & =\frac 1{2i}(Mz-\overline{Mz}) =\frac 1{2i}(Mz-M\overline{z}) =\frac 1{2i}(\frac{az+b}{cz+d}-\frac{a\bar z+b}{c\bar z+d}) \\ &=\frac 1{2i} \cdot\frac 1{|cz+d|^2}((az+b)(c\bar z+d) - (a\bar z+b)(cz+d)) \\ &=\frac 1{2i} \cdot\frac 1{|cz+d|^2}(ad-bc)(z-\bar z) \\ &=\frac {y_z}{|cz+d|^2}\ . \\[3mm] \text{So we finally need:}& \\ 0 &\overset{(!)}= -\frac{(cz+d)^2}y+2i\; c\;(cz+d)+\frac{(cz+d)(c\bar z+d)}y\ ,\text{ i.e.} \\ 0 &\overset{(!)}= -(cz+d)+2icy+(c\bar z+d)\ . \end{aligned} $$ Yes, this holds, since $y+\bar z = z$. (I did all the computations instead of only pointing out as in the first rows that there are two $Y$-values involved, namely $y_z$ and $y_{Mz}$, since very often this and such computations are highly depending on notation and details.)