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Solution

A chord through origin has equation $y=mx$. Its second intersection with $y^2=4x$ is found from $m^2x^2=4x$, giving $x=\dfrac{4}{m^2}, y=\dfrac{4}{m}$. Hence midpoint is $\left(\dfrac{2}{m^2},\dfrac{2}{m}\right)$, so its locus is $y^2=2x$. Let $P(x_1,y_1)$ lie on this curve. If a point $Q(x,y)$ divides $OP$ internally in ratio $3:1$, then $Q=\left(\dfrac{3x_1}{4},\dfrac{3y_1}{4}\right)$. Using $y_1^2=2x_1$, we get $\left(\dfrac{4y}{3}\right)^2=2\left(\dfrac{4x}{3}\right)$, which simplifies to $2y^2=3x$.