The Question is:
Let $(x^2+y^2)^3=(x^2-y^2)^2$ be a curve. Find the points on the curve where the normal line is parallel to y=0.
I have $\dfrac{dy}{dx}=\dfrac{-x(3x^4+6x^2y^2+3y^4-1)}{y(3x^4+6x^2y^2+3y^4+1)}$= a/0
as the slope of the normal line equals $\dfrac{-1}{(dy/dx)} $
How can I find the points on the curve?
First of all, your implicit differentiation went wrong. Here it goes: $$3(x^2+y^2)^2(2x+2yy')=2(x^2-y^2)(2x-2yy').$$ For our purpose it's better to take the derivative in respect to $y$, since we want $x'=0$, so $$3(x^2+y^2)^2(2xx'+2y)=2(x^2-y^2)(2xx'-2y).$$ Now if $x'=0$ we get $$3(x^2+y^2)^22y=2(x^2-y^2)(-2y).$$ We notice that $y=0$ is a solution, from which we get $x\in\{0,1,-1\}$.
If $y\neq0$, we have $$3(x^2+y^2)^2=-2(x^2-y^2).\qquad\qquad(*)$$ Squaring gives $$9(x^2+y^2)^4=4(x^2-y^2)^2.$$ Since $(x^2-y^2)^2=(x^2+y^2)^3$, we achieve $$x^2+y^2=4/9,$$ plugged in $(*)$ gives $$x^2-y^2=-8/27.$$ Adding the last two equations gives $$2x^2=\frac{4}{27}\iff x=\pm\sqrt{\frac{2}{27}}=\pm\frac{\sqrt2}{3\sqrt3}.$$ Now expand this fraction by $\sqrt3$ to get $$x=\pm\frac{\sqrt2\sqrt3}{3\sqrt3\sqrt3}=\pm\frac{\sqrt6}{9}.$$
Similarly, by subtracting both equations we get $y=\pm\sqrt{30}/9$.
Hence we have four more solutions, namely $$\frac19(\pm\sqrt6,\pm\sqrt{30}).$$
