By inspection, the curves intersect at (1,1) and (1,-1)
ellipse: 4x + 6yy' = 0
y' = -2x/3y
at (1,1) slope = -2/3
at (1,-1) slope = 2/3
semicubical parabola: 2yy' = 3x^2
y' = 3x/2y
at (1,1) slope = 3/2
at (1,-1) slope = -3/2
The slopes at the intersections are negative reciprocals; hence the curves are orthogonal.
two curves are orthogonal at a point of intersection of their tangents at that point cross at right angles. Show that the curves 2x^2+3y^2=5 and y^2=x^3 are orthogonal at (1,1) and (1,-1). Use parametric mode to draw the curves and to show the tangent lines
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