Determine Differentiable Function from Limit and Initial Value

To find the function $f(x)$, we begin by evaluating the given limit for each $x > 0$:
$\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1$

Since $x$ is held fixed as $t \to x$, this limit is in the indeterminate form $\frac{0}{0}$. Applying L'Hôpital's Rule with respect to the variable $t$:
$\lim_{t \to x} \frac{\frac{d}{dt}\left(t^2 f(x) - x^2 f(t)\right)}{\frac{d}{dt}(t - x)} = \lim_{t \to x} \frac{2t f(x) - x^2 f'(t)}{1}$

Evaluating the limit as $t \to x$:
$2x f(x) - x^2 f'(x) = 1$

Rearranging the terms gives a first-order linear differential equation in standard form:
$x^2 f'(x) - 2x f(x) = -1$
$f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2}$

To solve this linear differential equation, we determine the integrating factor $\mu(x)$:
$\mu(x) = e^{\int -\frac{2}{x} \, dx} = e^{-2 \ln x} = \frac{1}{x^2}$

Multiplying the differential equation by the integrating factor:
$\frac{1}{x^2} f'(x) - \frac{2}{x^3} f(x) = -\frac{1}{x^4}$
$\frac{d}{dx}\left(\frac{f(x)}{x^2}\right) = -\frac{1}{x^4}$

Integrating both sides with respect to $x$:
$\frac{f(x)}{x^2} = \int -x^{-4} \, dx = \frac{1}{3x^3} + C$
$f(x) = \frac{1}{3x} + Cx^2$

Using the initial condition $f(1) = 1$:
$1 = \frac{1}{3(1)} + C(1)^2 \implies 1 = \frac{1}{3} + C \implies C = \frac{2}{3}$

Substituting $C$ back into the expression for $f(x)$:
$f(x) = \frac{1}{3x} + \frac{2x^2}{3}$

Thus, the correct option is A.