showing that $R_k f =2if'+ky^{-1}f$ is a modular form of weight $k+2$ . 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
$$\frac{2ikcf(\tau)}{c\tau+d}+\frac{ky^{-1}f(\tau)}{(c\tau+d)^2}=ky^{-1}f(\tau).$$
 A: 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.)  
