if x-8y^3=0 express y as a function of x
$x - 8y^3 =0$
add $8y^3$ to both sides of the equation ...
$x = 8y^3$
divide both sides by $8$ ...
$\dfrac{x}{8} = y^3$
take the cube root of both sides ...
$\sqrt[3]{\dfrac{x}{8}} = \sqrt[3]{y^3}$
$\dfrac{\sqrt[3]{x}}{2} = y$
If we are having a disagreement about functions should we not have a clear notion what a function is?
That is a question foundations class.
Definition: Suppose that each of $A~\&~B$ is a nonempty set.
The statement that $f$ is a function from $A$ to $B$ means that:
$ \begin{align*}&\bullet~f\subseteq A\times B.\\&\bullet~\text{every }x\in A\text{ is the first term in exactly one pair in }f\\ \end{align*}$
Now if $A=B=\mathbb{R}$ it should be clear that $f=\{(x,x^2)\}$ is a function and $g=\{(x^2,x)\}$ is not!
Also $f=\{(x,x^3)\}~\&~g=\{(x^3,x)\}$ are both functions.