How deep is the liquid in a half-full hemisphere?

Assuming the spoon is a hemisphere with radius $R$,

let $x$ be the height from the bottom of the spoon, and let $h$ range from $0$ to $x$.

The radius $r$ of the circle at height $h$ satisfies $r^2=R^2-(R-h)^2=2hR-h^2$.

The volume of liquid in the spoon when it is filled to height $x$ is $$\int_0^x\pi r^2 dh=\int_0^x\pi(2hR-h^2)dh=\pi Rh^2-\frac13\pi h^3\mid_0^x=\pi Rx^2-\frac13\pi x^3.$$

(As a check, when the spoon is full, $x=R$ and the volume is $\frac23\pi R^3,$ that of a hemisphere.)

The spoon is half full when $\pi Rx^2-\frac13\pi x^3=\frac13\pi R^3;$ i.e., $3Rx^2-x^3=R^3;$

i.e., $a^3-3a^2+1=0$, where $a=x/R$.

The only physically meaningful solution of this cubic equation is $a\approx 65\%.$

There is actually an analytic solution to the problem, as shown below.

The volume of a spherical cap is the difference between those of two overlapping cones, one with a spherical bottom and the other with a flat bottom, i.e.

$$ V = \frac{2\pi}{3}r^2h - \frac{\pi}{3}(2rh-h^2)(r-h) =\frac{\pi}{3}(3rh^2-h^3)$$

Set $V$ to half of the semisphere volume $\frac{2\pi}{3}r^3$ to obtain,

$$\left(\frac rh \right)^3 - 3\frac rh+1=0$$

Compare with the identity $4\cos^3 x -3\cos x -\cos 3x=0$ and let $r/h = 2\cos x$ to obtain $x=40^\circ$.

Thus, the depth $h$ as a fraction of the radius $r$ Is

$$\frac hr = \frac{1}{2\cos40^\circ}$$

It makes things a bit simpler if we turn your measuring spoon upside down, and model it as the set of points $\{(x,y,z):x^2+y^2+z^2=1, z\ge 0\}$. The area of a cross-section at height $z$ is then $\pi(1-z^2)$, so the volume of the spoon between the planes $z=0$ and $z=h$ is

$$\pi\int_0^h(1-z^2)dz = \pi\left(h-\frac13h^3\right)$$

The volume of the hemisphere is $\frac23\pi$, and we want the integral to be equal to half this, i.e. $$\pi\left(h-\frac13h^3\right)=\frac{\pi}{3}$$ or $$h^3-3h+1=0$$ This cubic equation doesn't factorize nicely, so we ask Wolfram Alpha what it thinks. The relevant root is $h\approx 0.34730$. Remember that we turned the spoon upside down, so you should fill it to a height of $1-h=0.65270$, or $65.27\%$.