Answer
$$y'=\frac{(1+\ln x)^2-\ln x}{(1+\ln x)^2}$$
Work Step by Step
$$y=\frac{x\ln x}{1+\ln x}$$
Recall the following Derivative Rules: $$\frac{d}{dx}(\ln x)=\frac{1}{x}$$ $$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$ $$\frac{d}{dx}(uv)=v\frac{du}{dx}+u\frac{dv}{dx}$$
Therefore, we have $$y'=\frac{(x\ln x)'(1+\ln x)-(x\ln x)(1+\ln x)'}{(1+\ln x)^2}$$
We have $(x\ln x)'=(x)'\ln x+x(\ln x)'=\ln x+x\times\frac{1}{x}=\ln x+1$
$(1+\ln x)'=0+\frac{1}{x}=\frac{1}{x}$
$$y'=\frac{(\ln x+1)(1+\ln x)-(x\ln x)\times\frac{1}{x}}{(1+\ln x)^2}$$
$$y'=\frac{(1+\ln x)^2-\ln x}{(1+\ln x)^2}$$
