Answer
See explanations.
Work Step by Step
Step 1. Given the equation $y^2=x^3$, differentiate both sides with respect to $x$: $2yy;=3x^2$, which gives $y'=\frac{3x^2}{2y}$, and the slopes at $(1,\pm1)$ are $m_1=\frac{3}{2}$ and $m_2=-\frac{3}{2}$
Step 2. Similarly, given the equation $2x^2+3y^2=5$, differentiate both sides with respect to $x$: $4x+6yy'=0$, which gives $y'=-\frac{2x}{3y}$, and the slopes at $(1,\pm1)$ are $n_1=-\frac{2}{3}$ and $n_2=\frac{2}{3}$
Step 3. We can find that $m_1n_1=-1$ and $m_2n_2=-1$; thus the tangent lines of these two curves at the points $(1,\pm1)$ are perpendicular to each other at each point.
