is treating first order differential equations as results of chain rule valid? The entire treatise of differential equations seems obscure to me, it doesn't really have a foundation from which you can work from, as differentiation or integration do (that I know of--at least), like the difference quotient definition of a derivative, and infinite series definition of the Riemann integral.
But nonetheless, I tried to make sense of how you could work out a differential equation given only standard knowledge of differentiation/integration, and it follows as such:
The question asks to find $y$ such that:
$$ \frac{dy}{dx} = (2x + 3)\sqrt{y}$$
Now, this is a simple question, but if you notice that the RHS product could resemble a product of the chain rule, i.e. something of the form $[f(g(x))]' = g'(x)f'(g(x))$, now pick $g'(x) = \sqrt{y}$ and $f'(g(x) = 2x + 3$, then immediately you can solve $g(x) = \frac{2\sqrt{y^3}}{3} + c_0$, but then we have:
$$ \begin{align}
f'(g(x)) = f'\left(\frac{2\sqrt{y^3}}{3}\right) &= 2x + 3 \\
\therefore f(g(x)) = f\left(\frac{2\sqrt{y^3}}{3}\right) &= x^2 + 3x + c_1
\end{align}
$$
But this is not the answer, and resources don't seem to be abundant on this particular way of reasoning about ODEs, but I guess that the obstacle is to do with the fact that I am integrating with respect to $x$ in the second step on a function whose input is not necessarily dependent (?) on $x$, though I'm stuck.
If this method is aimless, then my question is to ask whether any related methods exist?
 A: There seem to be major conceptual flaws in your working. For instance, how do you go from:
$\displaystyle g'(x) = \sqrt{y}$
to $\displaystyle g(x) = \frac{2\sqrt{y^3}}{3} + c_0$ ?
You are integrating wrt $\displaystyle x$ on the LHS, but wrt $\displaystyle y$ on the RHS.
Basically, this equation is immediately solvable by separation of variables. What you must realise is that separation of variables is already based on the chain rule!
The whole point of ordinary differential equations is to try to get $\displaystyle y$ as a function of $\displaystyle x$, basically to find the function $h$ where $\displaystyle y=h(x)$.
So if you're given a simple ODE like $\displaystyle \frac{dy}{dx} = \frac{f(x)}{g(y)}$
You can then write $\displaystyle \frac{d}{dx}h(x) = \frac{f(x)}{g(y)}$
and proceed $\displaystyle h'(x) = \frac{f(x)}{g(y)}$ (from this point on, the prime notation denotes differentiation wrt $\displaystyle x$).
$\displaystyle h'(x)g(y) = f(x)$
$\displaystyle h'(x)g(h(x)) = f(x)$
At this point, by the (reverse of) chain rule, you should observe the LHS is the derivative of $\displaystyle G(h(x))$ wrt $\displaystyle x$, where $\displaystyle G'(x) = g(x)$ giving:
$\displaystyle h'(x) G'(h(x)) = f(x)$
$\displaystyle (G(h(x))'= f(x)$
$\displaystyle G(h(x)) = F(x) + c$
where $\displaystyle F'(x) = f(x)$.
and finally $\displaystyle y = h(x) = G^{-1}(F(x)+c)$.
assuming $\displaystyle G$ is invertible.
So separation of variables is already rooted in the application of chain rule. Our "normal" working by separating $\displaystyle y$ and $\displaystyle x$ to opposite sides is simply a shorthand representation that works.
A: The form of the chain rule you need to be using is that which you learned in multivariable calculus, an exact differential:
\begin{align}\tag{1}
\frac{\mathrm d}{\mathrm dx}\lambda(x,y)=\frac{\partial \lambda}{\partial y}\frac{\mathrm dy}{\mathrm dx}+\frac{\partial \lambda}{\partial x}.
\end{align}
You can read about it on the wiki page for chain rule or just google "multivariable chain rule". Rewriting your ODE gives
\begin{align}\tag{2}
\frac{\mathrm dy}{\mathrm dx}-(2x+3)\sqrt y=0,
\end{align}
which is similar in form to Eqn. (1), so we'll begin with the ansatz that Eqn. (2) is indeed an exact differential, so then it must be that
\begin{align}
1&=\frac{\partial \lambda}{\partial y},\ \ \ \ \text{and}\\
-(2x+3)\sqrt y&=\frac{\partial \lambda}{\partial x}.
\end{align}
If we integrate the first Eqn. we see that
\begin{align}
\lambda(x,y)=y+f(x),
\end{align}
where $f(x)$ is an arbitrary function picked up from integrating with respect to $y$. Taking a derivative with respect to $x$ and setting it equal to the other equation we get that
\begin{align}
\frac{\mathrm df}{\mathrm dx}=-(2x+3)\sqrt y.
\end{align}
But $f(x)$ is only a function of $x$, therefore the ansatz that Eqn. (2) is a perfect differential was wrong.
That being said, multiplying the entire equation by a integrating factor $m(x,y)$ sometimes makes your ODE a perfect differential, you can read about that by googling "exact differential equation integrating factor"
