Folium = Plot[y /. Solve[x^3 + y^3 == 3 x y, y], {x, -3, 3},
Epilog -> {PointSize[Large], Red, Point[{3/2, 3/2}]}]

Once you determine which of the three solution segments goes through $\{3/2, 3/2\}$ (it is the first) you find the derivative with respect to $x$. You can either take the derivative with respect to $x$ for the first solution, or merely cut and paste it as follows and substitute the $x$ value $3/2$:
Re@N@Simplify@
D[(2^(1/3) x)/(-x^3 + Sqrt[-4 x^3 + x^6])^(1/3)
+ (-x^3 + Sqrt[-4 x^3 + x^6])^(1/3)/2^(1/3), x] /. x -> 3/2
(* -1 *)
So the tangent line is $y = -x + b$, where $b$ ensures the line goes through $\{3/2, 3/2\}$. A simple Solve
shows that $b = 3$.
Thus the tangent line can be plotted as:
mylinePlot = Plot[-x + 3, {x, -3, 3}, PlotStyle -> Red];
Show[Folium, myLine]

The perpendicular has slope $-1$ divided by the slope of the tangent, and thus the slope is $+1$ and its intercept must ensure the perpendicular line go through $\{3/2, 3/2\}$. As before, a simple Solve
reveals $b = 0$:
myPerp = Plot[x, {x, -3, 3}, PlotStyle -> Green];
Show[Folium, myLine, myPerp, AspectRatio->1]
