Answer
$y=-2$
Work Step by Step
$y^2(y^2-4)=x^2(x^2-5)\\
(y^2)(2yy')+(y^2-4)(2yy')=(x^2)(2x)+(x^2-5)(2x)\\
2y^3y'+2y^3y'-8yy'=2x^3+2x^3-10x\\
4y^3y'-8yy'=4x^3-10x\\
2y'(2y^3-4y)=2(2x^3-5x)\\
y'=\frac{2x^3-5x}{2y^3-4y}$
Plug in $(0,-2)$ into $y′$ to find the gradient
$y'=\frac{2(0)^3-5(0)}{2(-2)^3-4(-2)}\\
y'=0$
Equation of the tangent line at $(0,-2)$
$y-(-2)=0(x-0)\\
y=-2$
