try the following:

CHAIN RULE: y=sin(x^sin(x)) y'=cos(x^sin(x))*x^sin(x)' = (I); now we forget the rest and only look at x^sin(x)

y=x^sin(x)

ln(y)=ln(x^sin(x))=sin(x)*ln(x) (as ln(a^b)=b*ln(a))

Now we do some differntiating

ln(y)'=y'/y, sin(x)*ln(x)'=sin(x)'*ln(x)+sin(x)*ln(x)'=cos(x)ln (x)+sin(x)/x

->y'/y=cos(x)ln(x)+sin(x)/x

Now the interesting part. We are not looking for ln(y)' but (as f(x)=y) we are looking fory', which we can find in the equation above. Hence, u solve the equation for y'

y'=y(cos(x)ln(x)+sin(x)/x)=x'sin(x)*(cos(x)ln(x)+sin(x)/x) (II)

Now plug (II) in in (I) and you have the solution

(x^sin(x)*cos(x^(sin(x))*(cos(x)ln(x)+sin(x))/x

Generally the point of logarithmic differentiation is the following:

f(x)=y=a^x

-> ln(y)=a*ln(x)

-> ln(y)'=(a*ln(x))'

-> y'/y = (a'ln(x) + a/x)

->y'=y*(a'ln(x) + a/x)