\[ x^{3}+y^{3}=9 ;(1,2) \] a. Use implicit differentiation to find $\frac{d y}{d x}$.

Step 1 ：We are given the equation \(x^{3}+y^{3}=9\) and the point (1,2). We are asked to find the derivative of \(y\) with respect to \(x\) at this point.

Step 2 ：We start by differentiating both sides of the equation with respect to \(x\). The derivative of \(x^3\) with respect to \(x\) is \(3x^2\). The derivative of \(y^3\) with respect to \(x\) is \(3y^2 \frac{dy}{dx}\), because \(y\) is a function of \(x\). The derivative of 9 with respect to \(x\) is 0. So, we get the equation \(3x^2 + 3y^2 \frac{dy}{dx} = 0\).

Step 3 ：Solving this equation for \(\frac{dy}{dx}\) gives \(\frac{dy}{dx} = -\frac{x^2}{y^2}\).

Step 4 ：We simplify this to get \(\frac{dy}{dx} = -\frac{x^2}{(9 - x^3)^{2/3}}\).

Step 5 ：We substitute the given point (1,2) into this equation to find the value of \(\frac{dy}{dx}\) at this point.

Step 6 ：Substituting \(x = 1\) and \(y = 2\) into the equation gives \(\frac{dy}{dx} = -0.25\).

Step 7 ：Final Answer: The derivative of \(y\) with respect to \(x\) at the point \((1,2)\) is \(\boxed{-0.25}\).