Answer
$y=Cx$
Work Step by Step
First we must find the slope of the tangent line that runs between $(x,y)$ and $(0,0)$.
$Slope=\frac{0−y}{0−x}=\frac{y}{x}$
Therefore, the derivative of the function f must be $\frac{y}{x}$.
$f′(x)=\frac{dy}{dx}=\frac{y}{x}$
$\frac{1}{y}dy=\frac{1}{x}dx$
$\ln{y}=\ln{x}+C_1$
$y=e^{\ln{x}+C_1}$
$y=e^{C_1}x$
$y=Cx$, where $C=e^{C_1}$
