The area of one of the two small right triangles formed by drawing lines through a point on the hypotenuse of a right triangle, parallel to the legs, dividing the triangle into a square and two smaller right triangles, is $\frac13$ times the area of the square. What is the ratio of the area of the other small right triangle to the area of the square?

Assume $a=BC, b=AC$ and $x=CD$ (the square's side). By $\triangle BEF\sim \triangle BAC$, $(a-x):a=x:b$, so $x=\frac{ab}{a+b}$.
The triangle $BEF$ has area $\frac12x(a-x)$, and the condition says $\frac12x(a-x)=\frac13x^2$, so $a-x=\frac23x$, i.e. $a=\frac53x$. Substituting into $x=\frac{ab}{a+b}$ gives $x(a+b)=ab$, so $\frac53x^2+xb=\frac53xb$, hence $b=\frac52x$. The other triangle $ADE$ has area $\frac12x(b-x)=\frac12x\cdot\frac32x=\frac34x^2$, so the ratio is $\boxed{\frac34}$.