I'm guessing that the expression should read something like
[tex]
r = \sin\theta\sin\phi[/tex]
where [tex]\theta[/tex] and [tex]\phi[/tex] are your polar angles. First, look in the x-y plane, where [tex]\theta=90^\circ[/tex]. This simplifies to
[tex]
r = \sin\phi.[/tex]
Now, apply [tex]r[/tex] to the equations linking polar and Cartesian coordinates, for [tex]\theta=90^\circ[/tex], as
[tex]
x = r\cos\phi[/tex]
[tex]
y = r\sin\phi.[/tex]
You'll see they come out to be
[tex]
x = \sin\phi\cos\phi[/tex]
[tex]
y = \sin^2\phi.[/tex]
Next, recall the double-angle formulas that [tex]\sin(2x) = 2\sin x\cos x[/tex] and [tex]\cos(2x) = 1 - 2 \sin^2 x[/tex]. I'll leave it as an exercise to substitute these back into the expressions for [tex]x[/tex] and [tex]y[/tex]. What you should notice is that your values should now look like
[tex]
x = A \sin(2\phi)[/tex]
and
[tex]
y = A \cos(2\phi) + B[/tex]
where [tex]A[/tex] and [tex]B[/tex] are numbers. You should recognize this as the parametric representation of a circle, at the coordinate [tex](0,B)[/tex].
Now, this doesn't exactly answer your question, but it should hopefully get you to visualize how the surface should be a sphere. Particularly, if you repeat this exercise in the y-z plane, where [tex]\phi=90^\circ[/tex], you should find find another case where the result is a circle, offset from the origin by some distance [tex]B[/tex].
I think that, once you've identified what [tex]B[/tex] is, you can look into the coordinate transformation of
[tex]
x' = x, \quad y' = y + B, \quad z' = z[/tex]
and then determine [tex]r' = \sqrt{x'^2 + y'^2 + z'^2}[/tex] which should be a constant. If this is the case, it shows that this surface is a sphere centered at [tex](0,B,0)[/tex].