Write the equation of the curve that passes through…

Write the equation of the curve that passes through $\left(1,\dfrac13\right)$ and for which the slope of the tangent line, at each point, is three times the slope of the radius vector at that point.

Explanation of Method

Translate the geometric condition into an ODE relating $y'$ and $y/x$ (the radius-vector slope), solve by separation of variables, and fit the constant to the given point.

Step 1: Set up the ODE

The radius-vector slope at $(x,y)$ is $y/x$, so the condition $y'=3\cdot\dfrac{y}{x}$ gives

\[\frac{dy}{y}=\frac{3\,dx}{x},\qquad x,y\neq0.\]

Step 2: Integrate

\[\ln|y|=3\ln|x|+C \quad\Longrightarrow\quad \boxed{\,y=Cx^3.}\]

Step 3: Apply the point condition

At $(1,\tfrac13)$: $\tfrac13=C\cdot1^3\Rightarrow C=\tfrac13$, giving

\[\boxed{\,y=\frac{x^3}{3}.}\]

Result

Final answer:

\[\boxed{\,y=\dfrac{x^3}{3}.}\]