Conservation fixes the total, resistance fixes the shares

At any junction,

$$\sum I_{\text{in}} = \sum I_{\text{out}}.$$

Charge is not created or stored at a node, so whatever arrives has to leave. That is one equation, and it constrains the sum.

To get the individual branch currents, use the fact that parallel branches share a potential difference, which the loop rule supplies:

$$I_k = \frac{\Delta V}{R_k}.$$

So $6$ A arriving at a junction between a $2\ \Omega$ branch and a $4\ \Omega$ branch divides as $4$ A and $2$ A, in inverse proportion to the resistances. Those still sum to $6$ A, which is why an even split of $3$ and $3$ passes the conservation check and fails the physics. Equal branches are the one case where the even split is right.

It is worth having the ratio in a usable form. For two branches sharing $\Delta V$,

$$\frac{I_1}{I_2} = \frac{R_2}{R_1},$$

which is the reverse of the ratio for a series voltage divider, where the larger resistance takes the larger share. Both follow from the same two facts, and the difference is which quantity the elements have in common.

1. Series: shared current, so $\Delta V = IR$ divides with the resistances.
2. Parallel: shared potential difference, so $I = \Delta V/R$ divides against them.

A sanity check that catches sign and ratio errors at once: the branch currents must sum to the arriving current, and each must be smaller than it.

Take one bulb in series feeding two bulbs in parallel, and let one of the parallel bulbs burn out. The untouched parallel bulb does not keep its old brightness, because removing a branch changes $R_{\text{eq}}$, which changes the battery current, which changes every potential difference in the circuit.

Trace it through:

1. Losing a parallel branch raises $R_{\text{eq}}$.
2. So the total current falls, and the series bulb, which carries all of it, dims.
3. The series bulb's drop therefore falls, leaving more potential difference across the parallel section.
4. So the surviving parallel bulb brightens.

One removal, and it dims one bulb while brightening another. Answering from the untouched branch alone is what misses it.

Every "what happens to the brightness" question is the same four steps, run before and after the change.

1. Compute $R_{\text{eq}}$ for the whole network.
2. Get the total current from $I = \varepsilon/R_{\text{eq}}$, using $\varepsilon - Ir$ if internal resistance is given.
3. Work outward: potential difference across each section, then the current in each branch.
4. Compare each bulb's $P = I^2R$ with what it had before.

It is slower than reasoning from the picture and it is right every time, including in the mixed circuits where the picture reasoning goes wrong. Once the two solved states are side by side, the brightness comparison reads straight off them.