Answer
$-1$, $\sqrt 3$
Work Step by Step
Step 1. Given the function $y^4=y^2-x^2$, differentiate both sides with respect to $x$ to get
$4y^3y'=2yy'-2x$ which gives $y'=\frac{x}{y-2y^3}$
Step 2. For point $(\frac{\sqrt 3}{4},\frac{\sqrt 3}{2})$, the slope $m_1=y'=\frac{\frac{\sqrt 3}{4}}{\frac{\sqrt 3}{2}-2(\frac{\sqrt 3}{2})^3}=\frac{\sqrt 3}{2\sqrt 3-3\sqrt 3}=-1$
Step 3. For point $(\frac{\sqrt 3}{4},\frac{1}{2})$, the slope $m_1=y'=\frac{\frac{\sqrt 3}{4}}{\frac{1}{2}-2(\frac{1}{2})^3}=\frac{\sqrt 3}{2-1}=\sqrt 3$
