For perfectly elastic collision,

total momentum before collision $=$ total momentum after collision

i.e. $mu+\frac{3m}{4}\times 0=m{v}_x+\frac{3m}{4}{v}_y$

$\Rightarrow 4u=4v_x+3v_y ...\left(\mathrm{i}\right)$

Coefficient of restitution,

$e=\frac{\text{Velocity of separation after collision} }{\text{Velocity of approach before collision} }=\frac{ \left(v_y-v_x\right)}{\left(u_x-u_y\right)}$

$\Rightarrow 1=\frac{v_y-v_x}{u-0}$   [$\because u_x=u$ and $u_y=0$]

$u=v_y-v_x ...\left(\mathrm{ii}\right)$

From equation (i) and (ii), we get

$v_x=\frac{u}{7}$ and $v_y=\frac{8u}{7}$

$\therefore \frac{{\lambda }_y}{{\lambda }_x}=\frac{{\displaystyle \frac{h}{{\displaystyle \frac{3}{4}}m{v}_y}}}{{\displaystyle \frac{h}{m\times v_x}}}=\frac{4}{3}\times \frac{v_x}{v_y}=\frac{4}{3}\frac{{\displaystyle \frac{u}{7}}}{{\displaystyle \frac{8u}{7}}}=\frac{1}{6}$