= Solution
Along the particle's worldline,
$$
x=u_xt,\qquad y=u_yt.
$$
The boost from $S$ to $S'$ gives
$$
x'=\gamma(x-Vt),\qquad
y'=y,\qquad
t'=\gamma\left(t-\frac{Vx}{c^2}\right).
$$
Therefore
$$
\frac{dt'}{dt}
=\gamma\left(1-\frac{Vu_x}{c^2}\right),
$$
and division of the transformed coordinate velocities by this factor gives the <relativistic velocity-addition formula>
$$
\boxed{
u'_x=\frac{u_x-V}{1-Vu_x/c^2},
\qquad
u'_y=\frac{u_y}{\gamma(1-Vu_x/c^2)},
\qquad
u'_z=0}.
$$
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