Step 1: Simplify the function first.
Use the identity $\text{sech}^{-1}(u)=\cosh^{-1}\!\left(\frac{1}{u}\right)$. Here $u=\frac{9}{9x^2+10}$, so $\frac{1}{u}=\frac{9x^2+10}{9}$. \[ y=\cosh^{-1}\!\left(\frac{9x^2+10}{9}\right)=\cosh^{-1}\!\left(x^2+\tfrac{10}{9}\right) \]
Step 2: Recall the cosh inverse derivative.
\[ \frac{d}{dx}\cosh^{-1}(u)=\frac{1}{\sqrt{u^2-1}}\cdot\frac{du}{dx} \]
Step 3: Differentiate the inside.
With $u=x^2+\frac{10}{9}$, $\frac{du}{dx}=2x$.
Step 4: Substitute.
\[ \frac{dy}{dx}=\frac{2x}{\sqrt{\left(x^2+\tfrac{10}{9}\right)^2-1}} \]
Step 5: Simplify the square root using difference of squares.
\[ \left(x^2+\tfrac{10}{9}\right)^2-1=\left(x^2+\tfrac{1}{9}\right)\left(x^2+\tfrac{19}{9}\right)=\frac{(9x^2+1)(9x^2+19)}{81} \]
Step 6: Take the root and finish.
The root brings out $\frac{1}{9}$, so \[ \frac{dy}{dx}=\frac{2x\cdot 9}{\sqrt{(9x^2+1)(9x^2+19)}}=\frac{18x}{\sqrt{(9x^2+19)(9x^2+1)}} \] \[ \boxed{\dfrac{18x}{\sqrt{(9x^2+19)(9x^2+1)}}} \]