I'll try to show my work below, hopefully you can read it:
y = sin(x^cos(x^2))
ln y = ln sin(x^cos(x^2))
d/dx(ln y) = d/dx(ln sin(x^cos(x^2)))
1/y * dy/dx = d/dx(ln sin(x^cos(x^2))) * d/dx(sin(x^cos(x^2))) * d/dx(x^cos(x^2)) * d/dx(cos(x^2)) * d/dx(x^2)
1/y * dy/dx = 1/sin(x^cos(x^2)) * cos(x^cos(x^2)) * cos(x^2)x^(cos(x^2)-1) * -sin(x^2) * 2x
dy/dx = 1/sin(x^cos(x^2)) * cos(x^cos(x^2)) * cos(x^2)x^(cos(x^2)-1) * -sin(x^2) * 2x * y
dy/dx = 1/sin(x^cos(x^2)) * cos(x^cos(x^2)) * cos(x^2)x^(cos(x^2)-1) * -sin(x^2) * 2x * sin(x^cos(x^2))
dy/dx = cos(x^cos(x^2)) * cos(x^2)* -sin(x^2) * 2x^(cos(x^2))
Ok, I see now. Taking the log of the equation as it stands doesn't help anything; if anything, it gets in the way. Logarithmic differentiation is used when taking a log simplifies the equation, not just because someone told you to!
What I did was first to take the inverse sine of each side:
y = sin(x^cos(x^2))
sin^(-1)(y) = x^cos(x^2)
Now we have a power, which the log will dismantle:
ln(sin^(-1)(y)) = ln(x^cos(x^2))
See what you can do starting there.
As a side comment, although you did the right things in most places, the notation in red is improper; you are really taking the derivative with respect to some intermediate function "u" in each case. It's incorrect to write the chain rule that way.
The main error you made is exactly what logarithmic differentiation is intended to make unnecessary, namely differentiating a power with the variable in both the base and the exponent. For a function of the form u^v, where u and v are functions of x, logarithmic differentiation yields this:
y = u^v
ln(y) = v ln(u)
y'/y = v*1/u*u' + v'*ln(u) = u'v/u + ln(u)v'
y' = (u^v)(u'v/u) + (u^v)ln(u)v'
y' =
vu^(v-1)u' + ln(u)v'u^v
This is a formula you could in principle memorize and apply to functions like
x^cos(x^2), but logarithmic differentiation lets you avoid that mess. Notice that what you wrote amounts to just the
first term of this formula, missing the fact that the exponent is not constant.