Step-by-step Solution:

1. Analyze the given limit expression. We are given the equation: $\mathop {\lim }\limits_{t \to x} {{{t^2}f(x) - {x^2}f(t)} \over {t - x}} = 1$ As $t \to x$, the numerator becomes $x^2f(x) - x^2f(x) = 0$ and the denominator becomes $x - x = 0$. This is an indeterminate form of type $\frac{0}{0}$. We can solve this limit by either using L'Hôpital's Rule or by rewriting the expression to relate it to the definition of a derivative.

2. Rewrite the limit to find the derivative. Let's add and subtract the term $x^2f(x)$ in the numerator: $\mathop {\lim }\limits_{t \to x} {{{t^2}f(x) - x^2f(x) + x^2f(x) - {x^2}f(t)} \over {t - x}} = 1$ Now, we can split the limit into two parts: $\mathop {\lim }\limits_{t \to x} \left( {{{t^2}f(x) - x^2f(x)} \over {t - x}} + {{x^2f(x) - {x^2}f(t)} \over {t - x}} \right) = 1$ Factor out common terms: $\mathop {\lim }\limits_{t \to x} \left( {f(x){{t^2 - x^2} \over {t - x}} - x^2{{f(t) - f(x)} \over {t - x}}} \right) = 1$ The term $t^2 - x^2$ can be factored as $(t-x)(t+x)$. $\mathop {\lim }\limits_{t \to x} \left( {f(x){{(t - x)(t + x)} \over {t - x}}} \right) - \mathop {\lim }\limits_{t \to x} \left( {x^2{{f(t) - f(x)} \over {t - x}}} \right) = 1$ The term $\frac{t-x}{t-x}$ cancels out. The second limit is the definition of the derivative $f'(x)$. $\mathop {\lim }\limits_{t \to x} \left( {f(x)(t + x)} \right) - x^2 \cdot f'(x) = 1$ Now, we can evaluate the limit by substituting $t=x$: $f(x)(x + x) - x^2 f'(x) = 1$ $2x f(x) - x^2 f'(x) = 1$

3. Formulate and solve the differential equation. The equation we derived is a first-order linear differential equation: $x^2 f'(x) - 2x f(x) = -1$ Let $y = f(x)$, so $y' = f'(x)$. $x^2 \frac{dy}{dx} - 2xy = -1$ Dividing by $x^2$ (since $x>0$), we get the standard form $\frac{dy}{dx} + P(x)y = Q(x)$: $\frac{dy}{dx} - \frac{2}{x} y = -\frac{1}{x^2}$ Here, $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{1}{x^2}$.

4. Find the integrating factor (I.F.). The integrating factor is given by $e^{\int P(x) dx}$. $I.F. = e^{\int -\frac{2}{x} dx} = e^{-2 \ln|x|} = e^{\ln(x^{-2})} = x^{-2} = \frac{1}{x^2}$ Since $x>0$, we can drop the absolute value.

5. Solve for the general solution. Multiply the standard form of the differential equation by the I.F.: $\frac{1}{x^2} \left( \frac{dy}{dx} - \frac{2}{x} y \right) = \frac{1}{x^2} \left( -\frac{1}{x^2} \right)$ $\frac{1}{x^2} \frac{dy}{dx} - \frac{2}{x^3} y = -\frac{1}{x^4}$ The left side is the derivative of $(y \cdot I.F.)$: $\frac{d}{dx} \left( y \cdot \frac{1}{x^2} \right) = -\frac{1}{x^4}$ Integrate both sides with respect to $x$: $\int \frac{d}{dx} \left( \frac{y}{x^2} \right) dx = \int -x^{-4} dx$ $\frac{y}{x^2} = -\frac{x^{-3}}{-3} + C$ $\frac{y}{x^2} = \frac{1}{3x^3} + C$ Solving for $y = f(x)$: $y = f(x) = x^2 \left( \frac{1}{3x^3} + C \right) = \frac{1}{3x} + Cx^2$

6. Use the initial condition to find the constant C. We are given that $f(1) = 1$. Substitute $x=1$ into the general solution: $f(1) = \frac{1}{3(1)} + C(1)^2 = 1$ $\frac{1}{3} + C = 1$ $C = 1 - \frac{1}{3} = \frac{2}{3}$

7. Write the final expression for f(x). Substitute the value of C back into the general solution: $f(x) = \frac{1}{3x} + \frac{2}{3}x^2$ This can also be written as: $f(x) = {1 \over {3x}} + {{2{x^2}} \over 3}$

8. Compare with the given options. The derived function $f(x) = {1 \over {3x}} + {{2{x^2}} \over 3}$ matches option A.

Alternative method using L'Hôpital's Rule: Differentiating numerator and denominator with respect to $t$ gives: $\mathop {\lim }\limits_{t \to x} {{d/dt(t^2f(x) - {x^2}f(t))} \over {d/dt(t - x)}} = \mathop {\lim }\limits_{t \to x} {{2tf(x) - x^2f'(t)} \over 1} = 2xf(x) - x^2f'(x)$ Setting this equal to 1 gives $2xf(x) - x^2f'(x) = 1$, which is the same differential equation obtained above, leading to the same solution.