My professor gave us an activity of, to me what feels like, a vague third derivative implicit differentiation. We were taught up to second, but now I am feeling lost.
I am tasked to find $\frac{d^3y}{dx^3}$, $x^2$+$y^2$=$a^2$.
I treated $a^2$ as a constant like what tutorial videos online did, and arrived at a different answer: $\frac{x}{27y^5}$. None of the choices match this in the question given to me, they all include a in them. Am I looking at $a^2$ differently here?
For general $a$ the answer should be
$$y'''(x) = \frac{-3a^2x}{|y|^{5}}.$$
To see this, we take the equation $x^2+y^2=a^2$ and differentiate it several times. The first, second and third derivatives give
$$2(x+yy') =0$$ $$ 2(1+ y'^2+ y y'')=0$$ $$6y'y'' +2yy'''=0$$ respectively, precisely because as you say, $a$ is a constant.
The first equation can be solved for $y'$ in terms of $y$. The second equation can be solved for $y''$ in terms of $y$ and $y'$ and hence, substituting the form of $y'$, this yields $y''$ in terms of just $y$. Finally the third equation can be solved for $y'''$ in terms of $y,y'$ and $y''$, and by substituting the previously found expressions of $y'$ and $y''$ this gives us a formula for $y'''$ in terms of $y$. It is given by
$$y'''(x) = \frac{-3x(x^2 + y^2)}{y^5}.$$
We can simplify this by using the fact that $x^2 + y^2=a^2$, so that
$$y'''(x) = \frac{-3xa^2}{y^5}.$$
This looks a lot like your answer, so perhaps you just made an error somewhere. Another possibility is that it has been specified somewhere that $a = 1/9$. In that case we would get exactly your answer, up to a minus sign.
EDIT: By the way, I just realized another possibility is that this was not meant as an implicit differentiation exercise (as is not completely clear from your question). Indeed the equation $x^2+y^2=a^2$ simply implies a formula for $y(x)$, namely
$$y(x) = \pm \sqrt{a^2-x^2}.$$
Perhaps the goal is simply to explicitly differentiate this expression three times.
