From Functions to Functions only.

Assuming $f(x)$ is both continuous and differentiable, let $y = x^2$ such that the above functional equation transforms into:

$f(x) + f(y) = \frac{y}{x}$

with the initial conditions at $x = 0, 1 \Rightarrow f(0) + f(0^2) = 0 \rightarrow f(0) = 0$ and $f(1) + f(1^2) = 1 \rightarrow f(1) = \frac{1}{2}.$ Differentiating the functional equation with respect to x and y each gives:

$f'(x) = -\frac{y}{x^2}$ (i) and $f'(y) = \frac{1}{x}$ (ii)

and equating (i) with (ii) yields:

$f'(x) = -\frac{y}{x^2} = -\frac{y \cdot f'(y)}{x} = \frac{A}{x}$ (iii)

and integrating (iii) produces:

$f(x) = A \cdot ln(x) + B$ (iv).

Because we have a logarithmic function, the initial condition $f(0) = 0$ cannot be applied. The other condition can be applied, which yields:

$A \cdot ln(1) + B = \frac{1}{2} \Rightarrow B = \frac{1}{2}$ , or $\boxed{f(x) = A \cdot ln(x) + \frac{1}{2}}.$ (v)

A final substitution of (v) into the original functional equation now produces:

$[A \cdot ln(x) + \frac{1}{2}] + [A \cdot ln(x^2) + \frac{1}{2}] \Rightarrow \boxed{3A \cdot ln(x) + 1 = x}$ ;

which does not hold true for ALL $x \in [0,1]$ . Thus, no such function exists!