I don't think there's many binomial coefficients in the expansion as in Frenkel-Lepowsky-Meurman's book (Prop 8.8.5). However, you can use the formal Taylor expansion [(8.8.10) in FLM]
$e^{-x \partial y} \delta(y/z) = \delta\left( \frac{y-x}{z} \right)$

from where your first identity becomes

(*) $\frac{y}{z} e^{-x \partial_y} \delta(y/z) = e^{x \partial_z} \delta(z/y)$

Now note that $\delta(y/z) = \delta(z/y)$ and both sides are power series in $x$. At $x = 0$ the RHS is simply $\delta(z/y)$ while the LHS is $\frac{y}{z} \delta(y/z) = \delta(y/z)$ by the first property of the $\delta$ function.

Taking derivatives with respect to $x$ on both sides $n$ times and putting $x=0$ you simply get $\partial^n_t \frac{y^{n+1}}{z^{n+1}} \delta (y/z) = \delta^{(n)} (z/y)$ where $t=y/z$ which is true for the same reasons.

To clean up a little and not use the issue of differentiating several times let us write the identity (*) more as the one you wrote (commuting some variables past its derivatives)
$e^{-x\partial_y} \frac{1}{z} \delta(y/z) = e^{x \partial_z} \frac{1}{y} \delta(z/y)$

Now note that we have the straightforward identities

$\frac{1}{z} \delta(y/z) = \frac{1}{y} \delta(y/z) = \frac{1}{y} \delta(z/y) = \frac{1}{z} \delta(z/y)$

And by direct computation (no binomial coefficients)

$\partial_y \frac{1}{z} \delta(y/z) = \sum_{n \in \mathbb{Z}} n y^{n-1} z^{-n-1} = - \sum_{n \in \mathbb{Z}} n y^{-n-1} z^{n-1}= -\partial_z \frac{1}{y} \delta(z/y)$

Finally passing $e^{x \partial_z}$ to the LHS your identity reads

$ e^{-x (\partial_y + \partial_z)} \frac{1}{z} \delta(y/z) = \frac{1}{y} \delta(z/y)$

and by the identity we just proved we have that $\partial_y + \partial_z$ acts by zero, hence the LHS is simply $\frac{1}{z} \delta(y/z)$ which trivially equals the RHS.

I am myself more used to the notation for $\delta(x-y) := \sum_{n \in \mathbb{Z}} x^n y^{-1-n}$ in which case the analogous formulas can be proved by using arguments as in Frenkel-Ben-Zvi's book by noting that $\delta$ is the difference of the expansions of $(x-y)^{-1}$ in different fields of the form $k( (x))((y))$ and $k((y))((x))$. From where identities like $(x-y) \delta(x-y)=0$ and $(\partial_x + \partial_y) \delta(x-y) = 0$ follow trivially by noting that both expansions of $1 = (x-y)/(x-y)$ agree.