= Solution
Now $m$ denotes the instantaneous <rest mass> of the rocket, including its fuel and stored <internal energy>, and $v$ is measured in a fixed <inertial frame>. Assume the exhaust is expelled backwards with speed $0<u<c$ in the rocket's instantaneous rest frame. By the <relativistic velocity-addition formula>, material emitted at time $\tau$ has inertial velocity
$$
w(\tau)=\frac{v(\tau)-u}{1-u v(\tau)/c^2}.
$$
It coasts at this velocity after emission.
The crucial change is that the expelled material's rest mass is not simply $-dm$: some of the rocket's <rest energy> becomes exhaust motion. Instead, use <conservation of relativistic energy>. Put $q(t)=m(t)\gamma(t)$, so the rocket's energy is $c^2q(t)$. The exhaust emitted during $d\tau$ carries energy $-c^2q'(\tau)d\tau$. For material moving at velocity $w$, the <relativistic energy-momentum relation> gives momentum $wE/c^2$, since $E=\gamma_{\rm ex}\mu c^2$ and $p=\gamma_{\rm ex}\mu w$. Hence the modified total <momentum> is
$$
P(t)=q(t)v(t)-\int_0^t w(\tau)q'(\tau)\,d\tau.
$$
Meanwhile the total relativistic energy is $c^2[q(t)-\int_0^tq'(\tau)d\tau]=c^2q(0)$. These expressions consistently account for both <energy> and <momentum> of the expelled material.
Differentiate the momentum formula. The <momentum conservation> law gives
$$
\boxed{\frac{d(m\gamma v)}{dt}
=\frac{v-u}{1-uv/c^2}\frac{d(m\gamma)}{dt}}.
$$
To simplify it, use
$$
qv'+(v-w)q'=0,\qquad
v-w=\frac{u}{\gamma^2(1-uv/c^2)},\qquad
q'=\gamma m'+m\gamma^3\frac{vv'}{c^2}.
$$
Multiplying by $\gamma^2(1-uv/c^2)$, the equation becomes
$$
m\gamma^3(1-uv/c^2)v'
+u\gamma m'+um\gamma^3vv'/c^2=0.
$$
The two terms containing $uv/c^2$ cancel. Dividing by $\gamma>0$ proves \b[the <relativistic rocket equation>]
$$
\boxed{m\gamma^2v'+um'=0}.
$$
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