Prove that the curve y=e^kxsin mx touches the curves

Prove that the curve $y=e^{kx}\sin mx$ touches the curves $y=e^{kx}$ and $y=-e^{kx}$ at all their common points.

Step 1: Common points

Since $e^{kx}>0$ always, $e^{kx}\sin mx=e^{kx}$ reduces to $\sin mx=1$, and $e^{kx}\sin mx=-e^{kx}$ reduces to $\sin mx=-1$. Either way, at a common point $x_0$, $\cos mx_{0}=0$.

Step 2: Slopes at a common point with $y=e^{kx}$

At $x_0$ with $\sin mx_0=1$, $\cos mx_0=0$: \[\bigl(e^{kx}\sin mx\bigr)'\big|_{x_0} = e^{kx_0}(k\sin mx_0+m\cos mx_0) = ke^{kx_0}, \qquad \bigl(e^{kx}\bigr)'\big|_{x_0}=ke^{kx_0}. \tag{1}\]

Step 3: Slopes at a common point with $y=-e^{kx}$

At $x_0$ with $\sin mx_0=-1$, $\cos mx_0=0$: \[\bigl(e^{kx}\sin mx\bigr)'\big|_{x_0} = e^{kx_0}(-k) = -ke^{kx_0}, \qquad \bigl(-e^{kx}\bigr)'\big|_{x_0}=-ke^{kx_0}. \tag{2}\]