Let z=-y then y'=-z'.
Let z=x^(-5/2).(e^7x).tan2x.
Taking logs of each side:
ln(z)=-(5/2)ln(x)+7x+ln(tan(2x)).
Differentiating:
z'/z=-5/2x+7+2sec^2(2x)/tan(2x)=-y'/(-y)=y'/y.
So y'=-x^(-5/2).(e^7x).tan(2x)(-5/2x+7+2sec^2(2x)/tan(2x)).
There are a few different ways of presenting the answer. For example, tan(2x) will cancel with the the third term in brackets containing tan(2x) in the denominator, and the x in the denominator of the first term in brackets could be used to alter the exponent of x to -7/2. The leading minus sign could be used to reverse the signs of the terms in brackets. However the equation is rearranged, it still looks complicated!