I read this as (ln²(sin(x)))cos(tan(x))/5xeˣ.
Let u=ln(sin(x)), du/dx=cos(x)/sin(x)=cot(x).
Let v=cos(tan(x)), dv/dx=-sin(tan(x))sec²(x).
Let w=5xeˣ, dw/dx=5eˣ+5xeˣ=5eˣ(1+x).
Let y=u²v/w, dy/dx=(w(u²dv/dx+2uvdu/dx)-u²vdw/dx)/w².
dy/dx=u(udv/dx+2vdu/dx)/w-(u²vdw/dx)/w².
dy/dx=u(-usin(tan(x))sec²(x)+2vcot(x))/w-(u²v(5eˣ(1+x))/w².
Now substitute for u, v, w:
dy/dx=ln(sin(x))(-ln(sin(x))sin(tan(x))sec²(x)+2cos(tan(x))cot(x))/5xeˣ-(ln²(sin(x))cos(tan(x))(eˣ(1+x))/5x²e²ˣ.