Let $$y = \log_e\left(\frac{1 - x^2}{1 + x^2}\right), -1 < x < 1$$. Then at $$x = \frac{1}{2}$$, the value of $$225(y' - y'')$$ is equal to

Simplify first: $$y = \ln(1-x^2) - \ln(1+x^2)$$.

First Derivative: $$y' = \frac{-2x}{1-x^2} - \frac{2x}{1+x^2} = \frac{-4x}{1-x^4}$$.

At $$x = 1/2$$: $$y' = \frac{-4(1/2)}{1 - (1/16)} = \frac{-2}{15/16} = -\frac{32}{15}$$.

Second Derivative: $$y'' = \frac{-4(1-x^4) - (-4x)(-4x^3)}{(1-x^4)^2} = \frac{-4 - 12x^4}{(1-x^4)^2}$$.

At $$x = 1/2$$: $$y'' = \frac{-4 - 12(1/16)}{(15/16)^2} = \frac{-4 - 3/4}{225/256} = \frac{-19/4}{225/256} = -\frac{19 \times 64}{225} = -\frac{1216}{225}$$.

$$225(y' - y'') = 225 \left( -\frac{32}{15} + \frac{1216}{225} \right) = 225 \left( -\frac{480}{225} + \frac{1216}{225} \right) = 736$$