The x-intercept of the tangent to a curve is equal to the ordinate of the point of contact. The equation of the curve through the point (1, 1) is

 $Y-y=m(X-x),ForX-interceptY=0$

$X=x-\frac{y}{m}$                    
Therefore, $x-\frac{y}{m}=y$ 
         
Or                $\frac{dy}{dx}=\frac{y}{x-y}$     
Put    $y=vxv+x\frac{dv}{dx}=\frac{v}{1-v}$
$x\frac{dv}{dx}=\frac{v}{1-v}-v=\frac{v-v+v^2}{1-v}$
$$\begin{gathered} \int \frac{1-v}{v^2}dv=\int \frac{dx}{x} \\ -\frac{1}{v}-lnv=lnx+C \\ -\frac{x}{y}-ln\frac{y}{x}=lnx+C\Rightarrow -\frac{x}{y}=lny+C \\ x=1,y=1\Rightarrow C=-1 \\ 1-\frac{x}{y}=lny\Rightarrow y=e.e^{-x/y} \\ {e}^{x/y}=\frac{e}{y}\Rightarrow y{e}^{x/y}=e \end{gathered}$$