The function $f(x)=x^2$ is not a parabola. It is merely a function.
If you write $y=x^2,$ then the set of points in the $x,y$-plane that satisfy that
equation is a parabola.
The "vertical line" test in a Cartesian plane tells you precisely whether a figure in the plane is a graph of a function of exactly one variable,
plotted using the horizontal axis (usually $x$)
to plot the input of the function and the vertical axis (usually $y$)
to plot the output of the function.
When you write $f(x,y)=\frac{x^2}9 - \frac{y^2}4,$
then certainly $f(x,y)$ is a function, but it's a function of two variables,
and you are not plotting its output using the vertical axis.
In fact, that equation does not even define any kind of curve in
the $x,y$-plane, because you've put no restrictions on what $x$ or $y$ could be.
Any values of $x$ and $y$ are valid input for this $f(x,y)$ and will
make the equation true, since you have defined the function that way.
If you write $$\frac{x^2}9 - \frac{y^2}4 = 0,$$
then you have written an equation describing a curve in the $x,y$-plane.
That curve is still not the graph of a function of $x$ whose output value is $y,$
and the vertical line test confirms that it is not.
Edit: I substituted new wording in a couple of places that I hope is a little more accurate than my previous use of the words "valid" and "relevant."
Moreover, I've attempted to make it clearer that the comments above apply only to
figures in a plane. You can define a figure in three-dimensional Cartesian space
by the points that satisfy $z=f(x,y)$ for some function $f$ of two variables,
and this figure will pass a "vertical line test" where the "vertical" lines are
parallel to the $z$-axis.
Thanks go to the commenters who pointed out these weaknesses in the original answer text.
The proof you give is more or less a sketch. It uses some not well-defined notions. A mathematical proof could run like this:
The linear function $f(x) = k x + d$ has direction vector $(1,k)$. The inverse $f^{-1}(x) = \frac{x-d}{k}$ only exists for $k \ne 0$ and has direction vector $(1,\frac{1}{k})$.
For the scalar product of the two direction vectors we get:
$(1,k) \cdot (1,\frac{1}{k}) = 1 + k \frac{1}{k} = 1 + 1 = 2$
which is non-zero, therefore the graphs are not perpendicular.
Best Answer
Everything you say is correct. $y^2 = 4ax$ is not a function of $x$, since a given (positive) value of $x$ corresponds to two possible values of $y$. But if you turn your head on its side, you see that the graph is the graph of a function of $y$ - any given value of $y$ corresponds to exactly one value of $x$.