Given equation of curve is
a2=x3
Differentiating w.r.t. x, we get
‌ 2ay ‌‌

dy

dx

=3x2
⇒‌

dy

dx

=‌

3x2

2ay

Slope of the tangent to the curve at (am2,am3) is
(‌

dy

dx

)(am2,am3)=‌

3(am2)2

2a(am3)

=‌

3a2m4

2a2m3

=‌

3m

2

Slope of normal at (am2,am3)
=‌

−1

‌ slope of the tangent at ‌(am2,am3)

=‌

−2

3m

Equation of the normal at (am2,am3) is
y−am3=‌

−2

3m

(x−am2)
⇒3my−3am4=−2x+2am2
⇒2x+3my−am2(2+3m2)=0