Functional Equation 2

Substitute $x = \frac{x-1}{x} \implies f(\frac{x-1}{x}) + f(\frac{1}{1-x}) = \frac{2x-1}{x}$ Call this A.

Now substitute $x = \frac{x-1}{x}$ into A to get $f(x) + f(\frac{x-1}{x}) = \frac{x-2}{x-1}$ and call this B.

Add the original identity and B and subtract A to get

$f(x) = \frac{x^3-x^2+1}{2x(x-1)}$

Now calculating $f(300)$ and rounding yields the answer of $\boxed{150.5}$

\(\begin{array} {} f(x) + f\left(\dfrac {x-1}x\right) = 1 + x & ...(1) & \small \color{blue} \text{Replace }x \text{ with }\dfrac {x-1}x. \\ f\left(\dfrac {x-1}x\right) + f\left(\dfrac 1{1-x}\right) = 2 - \dfrac 1x & ...(2) & \small \color{blue} \text{Replace }x \text{ with }\dfrac 1{1-x}. \\ f\left(\dfrac 1{1-x}\right) + f(x) = 1 + \dfrac 1{1-x} & ...(3) \end{array} \)

$\begin{aligned} (1)-(2)+(3): 2f(x) & = 1 + x - 2 + \frac 1x + 1 + \frac 1{1-x} \\ \implies f(x) & = \frac 12 \left(x + \frac 1x + \frac 1{1-x}\right) \\ f(300) & = \frac 12 \left(300 + \frac 1{300} - \frac 1{299}\right) \\ & \approx \boxed{150.0} \end{aligned}$

Alternative solution:

Substitute some values for $x$ for reasons I couldn't possibly justify other than they seemed likely to produce a useful result. Incidentally I used the Python sympy symbolic algebra module.

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>>> ex=Eq(f(x)+f((x-1)/x),1+x)
>>> ex.subs(x,300)
f(299/300) + f(300) == 301
>>> ex.subs(x,Rational(299,300))
f(-1/299) + f(299/300) == 599/300
>>> ex.subs(x,Rational(-1,299))
f(-1/299) + f(300) == 298/299

Then solve for the unknown function value.