a. Use implicit differentiation to find $\frac{d y}{d x}$ for $x^{2}+2 x-2 y+y^{2}=47$.
b. Use $\frac{d y}{d x}$ from part a. to find all possible slopes for the tangent line(s) when $y=1$.

Step 1 ：First, we need to find the derivative of the given equation. We can do this by applying the implicit differentiation to the equation \(x^{2}+2 x-2 y+y^{2}=47\).

Step 2 ：Taking the derivative of each term with respect to \(x\), we get \(2x + 2 - 2\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\).

Step 3 ：Rearranging the terms, we get \(\frac{dy}{dx}(2y - 2) = -2x - 2\).

Step 4 ：Dividing both sides by \(2y - 2\), we get \(\frac{dy}{dx} = \frac{-2x - 2}{2y - 2}\).

Step 5 ：Now, we need to find all possible slopes for the tangent line(s) when \(y=1\). We can do this by substituting \(y = 1\) into the derivative.

Step 6 ：Substituting \(y = 1\) into the derivative, we get \(\frac{dy}{dx} = \frac{-2x - 2}{2(1) - 2} = \frac{-2x - 2}{0}\).

Step 7 ：Since the denominator is zero, there are no possible slopes for the tangent line(s) when \(y=1\).

Step 8 ：So, the final answer is \(\boxed{\text{No possible slopes}}\).