Use the quotient rule when your function can be thought of as one function divided by another.
if #f(x) = (g(x))/(h(x))#
then #f'(x) = (g'(x)h(x)-g(x)h'(x))/(h^2(x))#
so if #f(x) = (3x)/sinx^2#
then #f'(x) = (3sinx^2 - 3x(2xcosx^2))/sin^2x^2#
so #f'(x) = (3sinx^2 - 6x^2cosx^2)/sin^2x^2#
Now I'm gonna assume we're working in degrees because #x=5# seems more like a value for degrees than radians, but I could be wrong. If I am the steps still apply it's just that my numbers will be off.
The derivative will give us the gradient, #m#, of the tangent, #y=mx+c#, at any given point #x#.
#m = (3sin25-150cos25)/(sin25)^2 = -134.7/0.1786#
#m = -754.2#
If you work with radians you should get around #-8520# as your gradient or something.
Now we have the equation
#y = -754.2x + c#
for which we still need the value for #c#, which we can obtain by substituting a point #(x,y)# which is on the tangent line.
We are already given a point #x=5#, so we should work out the corresponding #y# value:
#y = (3x)/(sinx^2) =15/sin25 = 35.49#
#:.#
#(x,y)=(5.000, 35.49)#
which we can substitute in to give us
#35.49 = -754.2 xx 5.000 + c#
#35.49 + 3771 = 3806 = c#
so
#y = -754.2x + 3806#
Obviously this will be a very different equation using radians, but the general principles are the same. I've also used all values to 4 significant figures.