By the Chain Rule (and knowledge of the derivative of inverse tangent function), the answer is: #dy/dx=\frac{1}{1+(3x)^{2})\cdot 3=\frac{3}{1+9x^{2}}#.
If you didn't happen to remember the derivative of the inverse tangent function, you could derive the answer by differentiation of both sides of the equation #tan(tan^{-1}(3x))=3x# with respect to #x# to get, by the Chain Rule, #sec^{2}(tan^{-1)(3x))\cdot \frac{d}{dx}(tan^{-1}(3x))=3# so that #\frac{d}{dx}(tan^{-1}(3x))=3cos^{2}(tan^{-1}(3x))#.
To simplify this, at this point, you could draw a right triangle and label one of the non-right angles #\tan^{-1}(3x)#. Since the tangent of the angle is the length of the opposite side divided by the length of the adjacent side, you could label the opposite side with "#3x#" and the adjacent side with "1". The Pythagorean Theorem would imply that the length of the hypotenuse is #\sqrt{1+9x^{2}}#. Since the cosine of the angle is the length of the adjacent side divided by the length of the hypotenuse, you'd get #\cos(\tan^{-1}(3x))=\frac{1}{\sqrt{1+9x^{2}}#.
Hence, #\frac{d}{dx}(tan^{-1}(3x))=3cos^{2}(tan^{-1}(3x))=\frac{3}{1+9x^{2}}#
