Let $f(y)=g(x)$ be an equation.
I remember my grade school teacher warning students against self-substitution in solving equations, which is the procedure described as follows:
Assuming $f(.)$ to be invertible, we solve for $y = f^{-1}(g(x))$. Then we substitute this expression for $y$ in the original equation:
$f(y)=g(x)$
$f(f^{-1}(g(x)))=g(x)$
$g(x)=g(x)$
Which is a tautology, since the substitution has introduced no new constraint and is thus a practically useless procedure.
Now, assume the equation can be expressed as $F(y,y)=f(y)=g(x)$. For instance, $F(a,b)$ could be $a^3+b$ and $F(y,y)=y^3+y$.
Questions:
1. Is it valid to apply the substitution $y = f^{-1}(g(x))$ to only one of the two $y$’s? Specifically, is it valid to say $F(y, f^{-1}(g(x)))=g(x)$?
2. If I wish to avoid the substitution, can $F(y, f^{-1}(g(x)))=g(x)$ be achieved through other means - through manipulations of the original equation $F(y,y)=g(x)$, i.e. by performing elementary operations on both sides?
I'm not really sure what your second question means; let me address the first.
The substitution your teacher warned you about is still valid - it won't lead to incorrect statements - it's just useless. So, you can do it, but there's no reason to.
As to the substitution you describe. First, let me describe an example. Suppose we're given the equation "$y+y=3x$." Then we have $F(a, b)=a+b$, $f(y)=y+y$ (so $f^{-1}(y)={y\over 2}$) and $g(x)=3x$. Then $y=f^{-1}(g(x))$, so this can be rewritten as $${3x\over 2}+y=3x.$$ I take this to be an example of what you're describing; if I'm wrong, please correct me. (Note that we can perform this style of substitution in many ways, e.g. to get ${3x\over 4}+{3y\over 2}=3x.$)
Is it valid? Sure, why wouldn't it be? Substituting an equal for an equal is always valid, that's what "equal" means.
Is it useful? Well, not really. In order to do this we need to already know $f^{-1}$, so the only possible use it could have is if $g$ were hard to invert, but $F$ and $g$ combined was easy to invert in the sense that the function $$G[f^{-1}(g(x)), g(x)]$$ were easy to compute - where $G$ is the "semi-inverse" of $F$: $G(a, b)=c$ if $F(a, c)=b$. Think of $G$ as taking two inputs, $a$ and $b$, and saying: "If $b$ were the output of an expression of the form $F(a, -)$, what would go in the blank?" The key point is that, because $$F(f^{-1}(g(x)), y)=g(x),$$ we have $$y=G[f^{-1}(g(x)), g(x)].$$ But I can't think of a case where this happens, so I'm going to say that this is probably not useful (except maybe in very contrived circumstances). I'd be happy to be corrected, though!