Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the reference frames are in the standard orientation with motion parallel to $x$ and $x'$ axes, $y=y'$.
So,
$$u'_y=\dfrac{dy'}{dt'}$$
where $t'=\gamma\left[ t-\dfrac{vx}{c^2} \right]$
$$u'_y=\dfrac{dy'}{d\left(\gamma\left[ t-\dfrac{vx}{c^2} \right]\right)}=\dfrac{dy'}{ \gamma dt-\dfrac{\gamma vdx}{c^2} }$$
$$u'_y =\dfrac{dy/dt}{ \gamma -\dfrac{\gamma v }{c^2} \dfrac{dx}{dt} }$$
$$\boxed{u'_y=\dfrac{u_y}{ \gamma \left[1 -\dfrac{v u_x}{c^2} \right] }}$$
By the same approach,
$$u_y=\dfrac{dy}{dt}$$
where $t =\gamma\left[ t'+\frac{v x'}{c^2} \right]$
$$u_y=\dfrac{dy}{\gamma dt'+\dfrac{v\gamma dx'}{c^2} }$$
$$u_y=\dfrac{dy'/dt'}{\gamma +\dfrac{v\gamma}{c^2}\dfrac{ dx'}{dt'} }$$
$$\boxed{u_y=\dfrac{u_y'}{\gamma \left[1+\dfrac{vu_x'}{c^2}\right] }}$$
$$\color{blue}{\bf [b]}$$
Let's assume that the Earth's frame is $\rm S$, and the rocket's frame is $\rm S'$, where $v_{\rm S'}=0.8 c$, and we are given that $u_y'=0.6 c$, and $u_x'=0$.
In frame $\rm S$, for Earth, the $x$-component of the projectile is given by
$$u_x=\dfrac{u_x'+v}{1+\dfrac{u_x'v}{c^2}}$$
Plug the known;
$$u_x=\dfrac{0+0.8 c}{1+\dfrac{0}{c^2}}= \color{blue}{0.8c}$$
In frame $\rm S$, for Earth, the $y$-component of the projectile is given by
$$u_y=\dfrac{u_y'}{\gamma \left[1+\dfrac{vu_x'}{c^2}\right] }$$
Plug the known;
$$u_y=\dfrac{0.6c}{\gamma \left[1+\dfrac{0}{c^2}\right] }=\dfrac{0.6 c}{\gamma}$$
$\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}=\left[1-\frac{0.8^2c^2}{c^2}\right]^{-1/2}=\frac{5}{3}$
$$u_y= \frac{3}{5} (0.6 c)= \color{blue}{0.36 c}$$
Therefore, the projectile’s speed in the earth’s reference frame is then
$$u=\sqrt{u_x^2+u_y^2}=\sqrt{0.8^2c^2+0.36^2c^2}$$
$$u=\color{red}{\bf 0.88}c$$
