= Solution
Inside the uniformly charged slab, its own <electric field> is $\rho(z+d/2)\hat{\mathbf z}/\epsilon_0$: it vanishes at the slab's midplane and has <derivative> $\rho/\epsilon_0$. Everywhere below the negative sheet, that sheet adds $+\sigma\hat{\mathbf z}/(2\epsilon_0)$. Therefore
$$
\boxed{\mathbf E(-d<z<0)=\frac{\rho(2z+d)+\sigma}{2\epsilon_0}\hat{\mathbf z}}.
$$
Its magnitude is the absolute value of this scalar coefficient. The field changes direction at
$$
\boxed{z_0=-\frac12\left(d+\frac\sigma\rho\right)},\qquad-d<z_0<-d/2.
$$
For $z<z_0$ it points downward; for $z>z_0$ it points upward, towards the negative sheet. Thus it is neither uniform nor directed upwards throughout the slab. The <derivative> and the field at the lower slab boundary also agree with <Gauss's law> and the previously found outside field.
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