We know that if the given equation were $\quad y = x^3 + ax^2 + bx$, $\quad$ then the derivative would be $3x^2 + 2ax + b$.
Since the given equation is different so the derivative will be: $$2(x^3 + ax^2 + bx)(3x^2 + 2ax + b) \implies 2y(3x^2 + 2ax + b)$$
What does: $$\frac{3x^2 + 2ax + b}{2y}$$ mean?
On
This is using implicit derivative. e.g. if we have:$$y^2=3x^2$$and we perform implicit derivative, we would get:$$2y\frac{dy}{dx}=6x$$$$\therefore \frac{dy}{dx}=\frac{6x}{2y}=\frac{3x}{y}$$ This is achieved by initially differentiating both sides w.r.t $x$ as follows:$$\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}\left(3x^2\right)=6x$$and then using the chain rule on the left-hand-side to get:$$\frac{d}{dy}\left(y^2\right)\frac{dy}{dx}=6x$$$$\therefore 2y\frac{dy}{dx}=6x$$
If you pretend that $y$ is a bona fide function of $x$ (it will be near most points on this curve), you can find the derivative of $y$ with respect to $x$ using implicit differentiation. Remember the chain rule: $\frac{d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$. That's what we'll do to find $\frac{dy}{dx} = y'(x)$ for the above equation. \begin{align*} \frac{d}{dx}\left[y(x)^2\right] &= \frac{d}{dx}\left[x^3 + ax^2 + bx\right]\\ \frac{d}{dx}\left[x^3 + ax^2 + bx\right] &= 3x^2 + 2ax + b\\ \frac{d}{dx}\left[y(x)^2\right] &= 2y(x)\cdot y'(x) = 2y\frac{dy}{dx}\\ \end{align*} Putting all this together, $$ 2y\frac{dy}{dx} = 3x^2 + 2ax + b. $$ Now, as long as $y$ isn't $0$, we can divide both sides by $2y$ to find $$ \frac{dy}{dx} = \frac{3x^2 + 2ax + b}{2y}. $$
What this equation is really telling you is that if you have a point $p = (x_0,y_0)$ such that $y_0^2 = x_0^3 + ax_0^2 + bx_0$ AND $y_0\neq 0$, then the slope of the curve defined by $y^2 = x^3 + ax^2 + bx$ at the point $p$ is $\frac{3x_0^2 + 2ax_0 + b}{2y_0}$.