Consider following differential equation
$$\frac{dy}{dx} = y^2(1+x^2)$$
Its solution is:
$$y(x) = \frac{-1}{x+\frac{1}{3}x^3+C}$$
As I read in one book, following must be satisfied:
$$\frac{dy(x)}{dx} = f(y(x))$$
And it is, plugging $y(x)$ to $y^2(1+x^2)$ will yield $y'(x)$, that is:
$$\frac{1+x^2}{(x+\frac{1}{3}x^3+C)^2}$$
Role of $\frac{dy(x)}{dx}$ is easy to grasp by intuition – $y(x)$ it's a function that matches slope "description" provided by $y^2(1+x^2)$, differentiating it will give the slope "description". But how to explain that $y(x)$ has to be substituted also on right side of $\frac{dy}{dx} = y^2(1+x^2)$?