Edit: I realise this is causing a lot of confusion, perhaps I've not been very clear as to what exactly I'm asking for, so here's a more clear summary.
I'm saying that if degree is positive integral power of the highest order derivative when the DE is expressible as a polynomial in derivatives then what is the degree of say, $ (y')^{1/3} = y''$?
According to me it should be undefined because it's not a polynomial in derivatives. But what I see most people do is raise both sides to the power $3$ then the new DE $y' = (y'')^3$ is a polynomial in derivatives with degree$3$, then they say the degree of the given DE is $3$.
All I'm saying is, isn't this kind of invalid because raising an equation to some power changes the equation? Not only that, I can raise it to different powers so that the resulting DE is a polynomial in derivatives, for eg: $(y')^{1/3} = y'' \Rightarrow (y')^3= (y'')^9$ which is a polynomial in derivatives, hence power of highest order derivative now is $9$ which is the degree of DE. You see, using that logic, we can get different degrees for same DE.
Firstly, I need someone to provide a proper definition of "degree" of an ODE.
The definition in my college textbook is, "Degree of an ODE is the highest power of the highest order derivative in the equation after clearing all radicals and fractions, if they exist."
Which is, I must say, impressively as garbage as a definition can get. For instance, according to that definition the order of $ \sin(y') + y = 5x^2 $ is $1$ since there are no fractions or radicals to be cleared, while the degree of that DE should be undefined.
Another definition, I see around on the internet is, "if the differential equation is a polynomial in derivatives, then the highest power of the highest order derivative is the degree of the ODE", cool, a little clearer and precise definition. But then they go on and say the degree of the DE $\sqrt{1 + y} = y'$ is $2$, that is, they basically square that DE on both sides to get a polynomial in derivatives but wait, wait, wait... I don't know how right I am about this, but I'm certain squaring an equation changes the equation, it's not the same equation anymore!!
Not only that, if you wanna play like that then the degree of the DE $(y'')^3 = y$ is $3$ while cubing the same DE will result in $(y'')^9 = y^3$ which is also a polynomial in derivatives, so it's degree is now $9$?
I also just went through first few pages of G.F. Simmons and I think he has totally skipped the topic of degree altogether. Nice.
The "degree" of a differential equation is, in an introductory level course, a teaching aid to be able to formulate a greater diversity of classification exercises. Which is useful when learning the "language" of differential equations, to get an intuition on what is syntactically correct and what is symbolic garbage. As there is no theoretical follow-up there is also no sharp definition, what is expected as answer depends on the textbook or course material, so all what you cited is likely correct in the context it was used.
In an advanced, specialized context the idea of an degree also makes sense in an algebraic treatment of purely algebraic differential equations (which means no sine permitted). It tells you how many (complex-valued) solution curves pass through any generic point of the state space. For that purpose, all branches of roots that occur in the differential equation are considered, which is the same as finding the minimal-degree polynomial that contains the given equation as one case. The degree of the DE is then the degree of the highest order derivative. $y'=y^3$ and $y'=x^9$ are both degree $1$.
For a practical demonstration, when you put such an algebraic differential equation (not DAE) in WolframAlpha, it gives the solution in that extended context, which may be confusing if the given differential equation is real and a real solution is expected.
In real calculus terms, it is as you observed, you only add spurious solutions by switching to this smallest polynomial DE. The degree does not really serve a generally useful purpose here.
As for the example $y'=y^{1/3}$, under the definition of "polynomial in the highest order derivative, everything else is coefficient functions", the degree is clearly $1$. Under the definition "all roots dissolved, minimal polynomial in all derivative orders", you get $y'^3=y$ as this polynomial and thus degree $3$. The degree $9$ expression is not minimal as polynomial.