In right triangle $DEF$, we have $\angle E = 90^\circ$ and $\tan D = 3\sin D$.  What is $\sin F$?
We start with a diagram:

[asy]

pair D,EE,F;

EE = (0,0);

F = (2*sqrt(2),0);

D = (0,1);

draw(D--EE--F--D);

draw(rightanglemark(F,EE,D,6));

label("$E$",EE,SW);

label("$F$",F,SE);

label("$D$",D,N);

//label("$9$",F/2,S);

[/asy]

First, we note that $\tan D = \frac{\sin D}{\cos D}$, so $\tan D = 3\sin D$ gives us $\frac{\sin D}{\cos D} = 3\sin D$, from which we have $\cos D = \frac13$.  Then, we note that $\sin F = \frac{DE}{DF} = \cos D = \boxed{\frac13}$.