The equation of the curve is,
$y^{4}=a x^{3}$
Differentiate w.r.t $x$, we get,
$4 y^{3} \frac{d y}{d x}=3 a x^{2}$
$\therefore \frac{ dy }{ dx }=\frac{3 ax ^{2}}{4 y ^{3}}$
$\therefore \frac{d y}{d x(a, a)}=\frac{3 a(a)^{2}}{4(a)^{3}}$
$\therefore \frac{d y}{d x(a, a)}=\frac{3(a)^{3}}{4(a)^{3}}$
$\therefore \frac{d y}{d x(a, a)}=\frac{3}{4}$
Thus, slope of tangent is $\frac{3}{4}$
Let $m _{1}=$ slope of normal
As tangent and normal are perpendicular to each other, we can write,
$\frac{3}{4} \times m_{1}=-1$
$\therefore m_{1}=\frac{-4}{3}$
Thus, equation of normal passing through $(a, a)$ is,
$y - y _{1}= m _{1}\left( x - x _{1}\right)$
$\therefore y - a =\frac{-4}{3}( x - a )$
$\therefore 3( y - a )=-4( x - a )$
$\therefore 3 y -3 a =-4 x +4 a$
$\therefore 4 x +3 y =7 a$