Why can't I take the 3rd root of both sides to simplify the implicit differentiation of $\tan^3\left(xy^2+y^{\:}\right)=x$? The task is to find the implicit derivative of $$\tan^3\left(xy^2+y^{\:}\right)=x$$
I can calculate
$$ \frac{dy}{dx}\mathrm{\:of\:}\tan ^3\left(xy^2+y\right)=x $$
as
$$\frac{1-3y^2\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)}{3\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
which agrees with the textbook, Calculus: Early Transcendentals (2016) by Anton, Bivens and Davis.
However, my initial idea was to take each side to the power of $\frac{1}{3}$. Yet the resulting calculation gives a different result:
$$
\frac{dy}{dx}\mathrm{\:of\:}\tan ^{\frac{3}{3}}\left(xy^2+y\right)=x^{\frac{1}{3}}
$$
now gives
$$\frac{1-3y^2x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)}{3x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
I was surprised by this result. I have the naive idea that, so long as I perform an operation on both sides of an equation, any manipulation is fair game.
I can see why I have this naive belief. After all, adding $2$ to both sides does not change the result:
$$\frac{dy}{dx}\mathrm{\:of\:}\tan ^3\left(xy^2+y\right)+2=x+2$$
is
$$ \frac{1-3y^2\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)}{3\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}$$
I can even continue with my manipulation by multiplying both sides by 2, still giving the same result:
$$\frac{dy}{dx}\mathrm{\:of\:}2\left(\tan ^3\left(xy^2+y\right)+2\right)=2\left(x+2\right)$$
is
$$\frac{1-3y^2\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)}{3\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}$$
But it seems my naive view of things has to change.
Some additional thoughts, although not very sophisticated, as I try to challenge my naive viewpoint, is that taking both sides to an even power or the reciprocal of an even power could be an issue.
I'm thinking that, after taking both sides to an even power, if I punted that equation on to someone else and told them only what I did and not what I started from, they wouldn't know whether the roots were originally positive or negative and could only guess at how to go back to the original equation.
And if I take both sides to the reciprocal of an even power, I could have negative values inside my radicals for some values of $x,y$, which is problematic for me (I only know of but don't use imaginary numbers).
Is that reasoning correct? Even if it is, I don't think I can apply the same reasoning to taking both sides to the reciprocal of an odd power. So I'm still mathematically naive about what is going on here.

It does yield the same result
$$
\frac{1-3y^2x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)}{3x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
we also have
$$
x^{1/3} = \tan\left(xy^2+y^{\:}\right)
$$
or
$$
x^{2/3} = \tan^2\left(xy^2+y^{\:}\right)
$$
insert into the above
$$
\frac{1-3y^2\tan^2\left(xy^2+y^{\:}\right)\sec ^2\left(xy^2+y\right)}{3\tan^2\left(xy^2+y^{\:}\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
which compared with your initial calculation
$$
\frac{1-3y^2\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)}{3\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
is consistent
