A differentiable function f satisfies the relation $f\left(xy\right)$ $=xf\left(y\right)+yf\left(x\right)\text{ for every }x,y\text{ in }{R}^{+}$.

If $f\left(1\right)=1,$ the find the function $f\left(x\right)$

Put $x=1,y=1,$ then $f\left(1\right)=0$

Now, $f\left(x\right)(1+h))=xf(1+h)+(1+h\left)f\right(x)$

$\begin{array}{lc}\Rightarrow & f(x+xh)-f\left(x\right)=xf(1+h)+hf\left(x\right)\\ \Rightarrow & f(x+xh)-f\left(x\right)\\ =& x\left(f\right(1+h)-f(1\left)\right)+hf\left(x\right)\\ \Rightarrow & \left(\frac{f(xh+x)-f\left(x\right)}{xh}\right)\cdot x\\ \Rightarrow & \underset{h\to 0}{lim} \left(\frac{f(xh+x)-f\left(x\right)}{xh}\right)x\\ \Rightarrow & \frac{x\left(f\right(1+h)-f(1\left)\right)}{h}+\frac{hf\left(x\right)}{h}\\ \Rightarrow & \underset{h\to 0}{lim} \left(\frac{f(1+h)-f\left(1\right)}{h}\right)x+f\left(x\right)\end{array}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x{f}^{\mathrm{\prime }}\left(x\right)=x{f}^{\mathrm{\prime }}\left(1\right)+f\left(x\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x{f}^{\mathrm{\prime }}\left(x\right)=x+f\left(x\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x\frac{dy}{dx}=x+y\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{dy}{dx}=1+\frac{y}{x}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{dy}{dx}+\left(-\frac{1}{x}\right)y=1\end{array}$

which is a linear differential equation.

$\therefore \mathrm{IF}=e^{-\int \frac{dx}{x}}=e^{-\log x}=\frac{1}{x}$

Hence, the solution is

$\begin{array}{r}\frac{y}{x}=\int \frac{dx}{x}+c\\ \frac{y}{x}=\log \left|x\right|+c\end{array}$

when $x=1,y=0,\text{ then }c=0$

Thus, the equation of the given curve is

$y=x\log \left|x\right|$