y = artanh(x/sqrt(1+x^2)) , find dy/dx

Some of these examiners quite like asking students to find the derivative of an inverse trig or hyperbolic function to try and catch someone off guard. The best way to approach these is to first multiply both sides by the regular function, in this case tanh to remove the nasty artanh function. Next we differentiate both sides of this equation implicitly, not forgetting to pop the dy/dx out of the y term we're differentiating, and then we inspect what we have. In this differentiated equation, we have a dy/dx term as just mentioned, and so we can see we're pretty close to the answer already. The only difficulty we might have in getting there is the weird sech squared y bit left over, but we know from our trig identities that sech squared has a simple relationship with tanh, of which you've already written down is equal to a function of x, so all we have to do is replace the sech squared with a tanh, then replace the tanh with our function of x, and suddenly we're left with an equation with only dy/dx and simple x terms, which we can easily rearrange to get our answer.