If  $y=\frac{x}{x^2+\frac{x}{x^2+\mathrm{......}\infty }},$ then  $\int \frac{(y-x^2)dx}{(x^2+y)(x+y^2)}=f\left(y\right)$ then  $100f\text{'}\left(4\right)$ is equal to 

$$\begin{gathered} \because y=\frac{x}{x^2+y}\Rightarrow y^2+x^{2}y-x=0 \\ \because \frac{dy}{dx}=\frac{-(2xy-1)}{(x^2+2y)}\Rightarrow \frac{dy}{dx}=\frac{-(2x\frac{x}{x^2+y}-1)}{(x^2+y+y)}=-\frac{-(2x^2-x^2-y)/x^{2}y}{(\frac{x}{y}+y)} \\ =\frac{-y(x^2-y)}{(x^2+y)(x+y^2)}\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{(y-x^2)}{(x^2+y)(x+y^2)} \\ =\frac{d}{dx}\left(\ln y\right)=\frac{(y-x^2)}{(x^2+y)(x+y^2)}\Rightarrow \ln y+c\int \frac{(y-x^2)dx}{(x^2+y)(x+y^2)} \end{gathered}$$