4-momentum and an $y$ component of momentum I have 2 coordinate systems which move along $x,x'$ axis. 
I have derived a Lorentz transformation for an $x$ component of momentum, which is one part of an 4-momentum vector $p_\mu$. This is my derivation:
$$
\scriptsize
\begin{split}
p_x &= mv_x \gamma(v_x)\\
p_x &= \frac{m (v_x'+u)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{1 - \left(v_x' + u \right)^2 / c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2}} \\
p_x &= \frac{m (v_x'+u) \left( 1+ v_x' \frac{u}{c^2} \right)}{\left(1+v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\
p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ v_x' \frac{u}{c^2} \right)^2 - \left(v_x' + u \right)^2 \right] / c^2 }} \\
p_x &= \frac{m (v_x'+u)}{\sqrt{\left[c^2 \left( 1+ 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} \right) - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + 2 v_x'u + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - 2 v_x' u - u^2 \right] / c^2 }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{\left[c^2 + v_x'^2 \frac{u^2}{c^2} - v_x'^2 - u^2 \right] / c^2 }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{1 + v_x'^2 \frac{u^2}{c^4} - \frac{v_x'^2}{c^2} - \frac{u^2}{c^2} }} \\
p_x &= \frac{mv_x'+mu}{\sqrt{\left(1 - \frac{u^2}{c^2}\right) \left(1-\frac{v_x'^2}{c^2} \right)}} \\
p_x &= \gamma \left[mv_x' \gamma(v_x') + mu \gamma(v_x') \right] \\
p_x &= \gamma \left[mv_x' \gamma(v_x') + \frac{mc^2 \gamma(v_x') u}{c^2} \right] \\
p_x &= \gamma \left[p_x' + \frac{W'}{c^2} u\right] 
\end{split}
$$
I tried to derive Lorentz transformation for momentum also in $y$ direction, but i can't seem to get relation $p_y=p_y'$ because in the end i can't get rid of $2v_x'\frac{u}{c^2}$ and $\frac{v_y'^2}{c^2}$. Here is my attempt.
$$
\scriptsize
\begin{split}
p_y &= m v_y \gamma(v_y)\\
p_y &= \frac{m v_y'}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{1 - v_y'^2/c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2}}\\
p_y &= \frac{m v_y' \left( 1 + v_x' \frac{u}{c^2} \right)}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + v_x' \frac{u}{c^2} \right)^2 - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2\left( 1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4}\right) - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{\left[c^2 + 2 v_x' u + v_x'^2 \frac{u^2}{c^2} - v_y'^2\right]/c^2}}\\
p_y &= \frac{m v_y'}{\gamma \sqrt{1 + 2 v_x' \frac{u}{c^2} + v_x'^2 \frac{u^2}{c^4} - \frac{v_y'^2}{c^2}}}\\
\end{split}
$$
This is where it ends for me and I would need someone to point me the way and show me, how i can i get $p_y = p_y'$? I know I am very close.

Well this is how $p_y$ part of a four-momentum is put together. 
\begin{equation}
\scriptsize
\begin{split}
p &= m v \gamma(v)\\
&\Downarrow\\
p_y &= m v_y \gamma(v) = m v_y \gamma \left( \sqrt{v_x^2 + v_y^2 + v_z^2}\right) = m v_y \gamma \left( \sqrt{v_x^2 + 0 + 0}\right) = m v_y \gamma(v_x) =\\
&= \frac{m v_y'}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{1 - \frac{\left(v_y' + u\right)^2}{c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2}}} = \frac{mv_y'}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{\frac{c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2 - \left(v_x' + u\right)^2}{c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2}}}=\\
&= \frac{mv_y' \left(1 + v_x' \frac{u}{c^2}\right)}{\gamma \left(1 + v_x' \frac{u}{c^2}\right) \sqrt{\left[c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2 - \left(v_x' + u\right)^2\right] / c^2}} = \frac{mv_y'}{\gamma \sqrt{\left[c^2 \left(1 + v_x' \frac{u}{c^2}\right)^2 - \left(v_x' + u\right)^2\right] / c^2}}=\\
&= \frac{mv_y'}{\gamma \sqrt{\left[c^2 \left(1 + 2 v_x' \frac{u}{c^2} + {v_x'}^2 \frac{u^2}{c^4}\right) - {v_x'}^2 - 2 {v_x'}u - u^2\right] / c^2}}=\\
& = \frac{mv_y'}{\gamma \sqrt{\left[c^2 + 2 v_x' u + {v_x'}^2 \frac{u^2}{c^2} - {v_x'}^2 - 2 {v_x'}u - u^2\right] / c^2}}= \frac{mv_y'}{\gamma \sqrt{\left[c^2 + {v_x'}^2 \frac{u^2}{c^2} - {v_x'}^2 - u^2\right] / c^2}}=\\
& = \frac{mv_y'}{\gamma \sqrt{1 + {v_x'}^2 \frac{u^2}{c^4} - \frac{{v_x'}^2}{c^2} - \frac{u^2}{c^2}}}= \frac{mv_y'}{\gamma \sqrt{\left(1 - \frac{u^2}{c^4}\right)  \left(1-\frac{{v_x'}^2}{c^2}\right)}}=  mv_y' \gamma(v_x')\\
\end{split}
\end{equation}
In our case $v_x' = v'$ and we can modify last part of this big equation so that we get:
\begin{equation}
\scriptsize
\begin{split}
p_y= mv_y' \gamma(v')\\
\end{split}
\end{equation}
Now we can see Lorentz tr. and reverse Lorentz tr. which are:
\begin{equation}
\scriptsize
\begin{split}
&\boxed{p_y=p_y'} ~~~\boxed{p_y'=p_y}\\
\end{split}
\end{equation}
