Solution (a) We use implicit differentiation: $$\begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{d}{dx}( e^{y}\cos x) &= \dfrac{d}{dx}(x+1) \\ e^{y}\cdot \dfrac{d}{dx}( \cos x) +\left( \dfrac{d}{dx} e^{y}\right) \cdot \cos x &= 1 & {\color{#0066A7}{\hbox{Use the Product Rule.}}} \\ e^{y}( -\sin x) +e^{y}y^\prime \cdot \cos x &= 1 \\ ( e^{y}\cos x) y^\prime &= 1+e^{y}\sin x \\ y^\prime &= \dfrac{1+e^{y}\sin x}{e^{y}\cos x} \end{array} \end{eqnarray*}$$

(b) At the point $(0,0)$, the derivative $y^\prime$ is $\dfrac{1+e^{0}\sin 0}{e^{0}\cos 0}=\dfrac{1+1\cdot 0}{1\cdot 1}=1$, so the slope of the tangent line to the graph at $(0,0)$ is $1$. An equation of the tangent line to the graph at the point $(0,0)$ is $y=x.$