unexpected case of gradient of inverse function

Question

if $y=x^3+3x+2$ is the original equation then, $$\frac{dy}{dx} = 3x^2+3$$ So the gradient of the inverse function atc $x=2$ should be $$\frac{dx}{dy}= \frac{1}{3x^2+3}$$ This gives me answer of $\frac1 {15}$ but when i graph the function it is clearly $\frac13$. Is there some situation when this rule doesn't apply or an error in my logic?

Answer 1

I believe that when you're looking at graph of the inverse function $x(y)$ you're looking at $y=2$, that is $x=0$. In this point you should actualy have the gradient to be equal to $\frac13$.