Answer
point $(-3,2)$, $m_1=\frac{-27}{8}$
point $(3,2)$, $m_2=\frac{27}{8}$
point $(-3,-2)$, $m_3=\frac{27}{8}$
point $(3,-2)$, $m_4=\frac{27}{-8}$
Work Step by Step
Step 1. Given the equation $y^4-4y^2=x^4-9x^2$, differentiate both sides with respect to $x$: $4y^3y'-8yy'=4x^3-18x$ which gives $y'=\frac{2x^3-9x}{2y^3-4y}$,
Step 2. the slope of the tangent line at point $(-3,2)$ is $m_1=y'=\frac{2(-3)^3-9(-3)}{2(2)^3-4(2)}=\frac{-54+27}{16-8}=\frac{-27}{8}$
Step 3. the slope of the tangent line at point $(3,2)$ is $m_2=y'=\frac{2(3)^3-9(3)}{2(2)^3-4(2)}=\frac{54-27}{16-8}=\frac{27}{8}$
Step 4. the slope of the tangent line at point $(-3,-2)$ is $m_3=y'=\frac{2(-3)^3-9(-3)}{2(-2)^3-4(-2)}=\frac{-54+27}{-16+8}=\frac{27}{8}$
Step 5. the slope of the tangent line at point $(3,-2)$ is $m_4=y'=\frac{2(3)^3-9(3)}{2(-2)^3-4(-2)}=\frac{54-27}{-16+8}=\frac{27}{-8}$
