Step 1: Identify $u$ and $v$ for the quotient rule. Given $y = \frac{x^2 + 3}{x^3 + 4}$. Let $u = x^2 + 3$ and $v = x^3 + 4$. Step 2: Find the derivatives of $u$ and $v$ with respect to $x$. $$ \frac{du}{dx} = \frac{d}{dx}(x^2 + 3) = 2x $$ $$ \frac{dv}{dx} = \frac{d}{dx}(x^3 + 4) = 3x^2 $$ Step 3: Apply the quotient rule. The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$. Substitute the expressions for $u, v, \frac{du}{dx}, \frac{dv}{dx}$: $$ \frac{dy}{dx} = \frac{(x^3 + 4)(2x) - (x^2 + 3)(3x^2)}{(x^3 + 4)^2} $$ Step 4: Simplify the expression for $\frac{dy}{dx}$. Expand the numerator: $$ \frac{dy}{dx} = \frac{2x^4 + 8x - (3x^4 + 9x^2)}{(x^3 + 4)^2} $$ $$ \frac{dy}{dx} = \frac{2x^4 + 8x - 3x^4 - 9x^2}{(x^3 + 4)^2} $$ Combine like terms in the numerator: $$ \frac{dy}{dx} = \frac{-x^4 - 9x^2 + 8x}{(x^3 + 4)^2} $$ Step 5: Evaluate $\frac{dy}{dx}$ at $x=1$. Substitute $x=1$ into the derivative expression: $$ \frac{dy}{dx}\Big|_{x=1} = \frac{-(1)^4 - 9(1)^2 + 8(1)}{((1)^3 + 4)^2} $$ $$ = \frac{-1 - 9 + 8}{(1 + 4)^2} $$ $$ = \frac{-10 + 8}{(5)^2} $$ $$ = \frac{-2}{25} $$ The value of $\frac{dy}{dx}$ at $x=1$ is: $$\boxed{\frac{-2}{25}}$$