If $a\left(4+x^2\right)=x$ and $\mathrm{y}-x^3=a^2$ then $\frac{\mathrm{dy}}{\mathrm{d} x}$ at $x=1$ is ...

From $a(4+x^2) = x$, we obtain $a = \frac{x}{4+x^2}$. Substituting into $y - x^3 = a^2$ yields $y = x^3 + \frac{x^2}{(4+x^2)^2}$.

Differentiating gives $\frac{dy}{dx} = 3x^2 + \frac{d}{dx}\left(\frac{x^2}{(4+x^2)^2}\right)$. Using the quotient rule:

$\frac{d}{dx}\left(\frac{x^2}{(4+x^2)^2}\right) = \frac{2x(4+x^2)^2 - x^2 \cdot 4x(4+x^2)}{(4+x^2)^4} = \frac{2x(4+x^2)[(4+x^2) - 2x^2]}{(4+x^2)^4} = \frac{2x(4-x^2)}{(4+x^2)^3}$.

Therefore, $\frac{dy}{dx} = 3x^2 + \frac{2x(4-x^2)}{(4+x^2)^3}$.

Evaluating at $x=1$: $\left.\frac{dy}{dx}\right|_{x=1} = 3 + \frac{2(4-1)}{(4+1)^3} = 3 + \frac{6}{125} = \frac{381}{125}$.

Final answer: $\boxed{\frac{381}{125}}$