Use implicit differentiation to justify the vertical asymptote (x = -3) for the curve

The derivative of the y function represents the slope of the function on each point. y is a function of x, if we differentiate our curve, we're going to have
[tex]\begin{gathered} \frac{d}{dx}(y^2)=\frac{d}{dx}(x^3+3x^2) \\ 2y\frac{dy}{dx}=3x^2+6x \\ \frac{dy}{dx}=\frac{3x^2+6x}{2y} \end{gathered}[/tex]The derivative at x = -3 diverges(when x = -3, y = 0), therefore, we have a vertical asymptote.