I can't solve it

I don't think you need to use L'hospital's Rule here.

This limit has the indeterminate form 0 * -oo.

sin x ln x = ln x / (1/sin x)

take derivatives:
1/x / (-1/sin^2 x * cos x)
= (1/x) / - cos x / sin^2 x
= -(1/x) sin^2 x / cos x
= -sin x / x * sin x/cos x

sin x / x --> 1

limit = 0.

you can always avoid using lhospital
anyway here, remember that sin(x) <= x for x > 0,

|sin(x) ln(x)| <= |x ln(x)|

now, since (x ln(x)) -> 0 when x goes to 0 you can write sin(x) ln(x) -> 0 when x goes to 0