The equation of the tangent to the curve $y=1-e^{x / 2}$ at the point of intersection with the y-axis, is

$x+2 y=0$

We have,

$y=1-e^{x / 2} \Rightarrow \frac{d y}{d x}=-\frac{1}{2} e^{x / 2}$

The curve $y=1-e^{x / 2}$ meets y-axis at (0, 0)

∴  $\left(\frac{d y}{d x}\right)_{(0,0)}=-\frac{1}{2}$

The equation of the tangent at (0, 0) is

$y-0=-\frac{1}{2}(x-0) \Rightarrow x+2 y=0$