How to solve ode of the form $ a_n(y')^n + a_{n-1}(y')^{n-1} + \cdots + a_1 y' + a_0 =0 $

Well. As pointed out in the comments by Seth.

Let us consider an example

\begin{align} (y')^2+2y'+1=0 \end{align} then it follows $y' = -1$ so $y = -t+C$.

So, in general, you will get $y'=$ const. Then $y=\text{const}. t+C$.

TLDR. All continuously differentiable solutions of the ODE are of the form $y(t) = c + tz$ where $z$ is a (possibly complex) root of the corresponding polynomial.

Let $y:[0,T)\rightarrow\mathbb{R}$ where $T\leq\infty$ be a continuously differentiable function satisfying the ODE $$ a_{n}(y^{\prime}(t))^{n}+\cdots+a_{1}y^{\prime}(t)+a_{0}=0. $$ By the fundamental theorem of algebra, if $a_{n}\neq0$, the corresponding polynomial $$ a_{n}r^{n}+\cdots+a_{1}r+a_{0} $$ has $n$ complex roots, call them $z_{1},\ldots,z_{n}$. We can re-express the ODE as $$ \left(y^{\prime}(t)-z_{1}\right)\cdots\left(y^{\prime}(t)-z_{n}\right)=0. $$ By the above, $y^{\prime}(t)=z_{k(t)}$ where $k(t)\in\{1,\ldots,n\}$ for all $t$. In fact, $t\mapsto z_{k(t)}$ must be constant since otherwise continuous differentiability is violated. We conclude that $y(t)=c+tz_{j}$ for some $j$.