= Solution
Use <left Grassmann derivatives> and define the <supersymmetric covariant derivatives> $D_\pm=\partial_{\theta^\pm}-i\theta^\pm\partial_\pm$. Multiplication in the indicated order gives exactly the printed differential operator:
$$
iD_-D_+=i\partial_{\theta^-}\partial_{\theta^+}+\theta^-\partial_{\theta^+}\partial_- -\theta^+\partial_{\theta^-}\partial_+ -i\theta^-\theta^+\partial_-\partial_+.
$$
Thus the equation is $iD_-D_+\Phi=W(\Phi)$ in <two-dimensional N=(1,1) superspace>.
Under a <Lorentz boost>, take $x^\pm\mapsto e^{\pm\omega}x^\pm$ and $\theta^\pm\mapsto e^{\pm\omega/2}\theta^\pm$. The derivatives $\partial_\pm$ have boost weights $\mp1$, and $D_\pm$ have weights $\mp1/2$. Consequently $D_-D_+$ has weight zero, and is a <Lorentz scalar>. Both $\Phi$ and the ordinary function $W(\Phi)$ are scalar <superfields>, so the equation is <Lorentz invariant>.
For <supersymmetry>, $\{\partial_\theta,\theta\}=1$ and anticommutation of distinct odd coordinates give
$$
\{D_s,Q_t\}=0\quad(s,t\in\{+,-\}),\qquad Q_\pm^2=i\partial_\pm,\qquad \{Q_+,Q_-\}=0.
$$
Moving either <supercharge> through the two odd derivatives introduces two minus signs, hence $[Q_t,D_-D_+]=0$. If $\delta\Phi=\epsilon^+Q_+\Phi+\epsilon^-Q_-\Phi$, with odd constant parameters, the <chain rule> for an even scalar <superfield> gives
$$
\delta\bigl(iD_-D_+\Phi-W(\Phi)\bigr)=\sum_{t=\pm}\epsilon^tQ_t\bigl(iD_-D_+\Phi-W(\Phi)\bigr).
$$
Therefore a solution is mapped to a solution, proving <supersymmetry> invariance.
To compute the <supersymmetric Liouville equation>, order the odd coordinates as $\theta^-\theta^+$ and anticommute the fermionic component fields with both coordinates. In particular, $\partial_{\theta^+}(\theta^-\theta^+)=-\theta^-$. Direct application of the operator gives
$$
iD_-D_+\Phi=F+i\theta^-\partial_-\psi_- -i\theta^+\partial_+\psi_+ -i\theta^-\theta^+\partial_-\partial_+\phi.
$$
Let $N=\Phi-\phi$. Only the two products of its degree-one terms survive in $N^2$; for example, $(i\theta^-\psi_+)(i\theta^+\psi_-)=\theta^-\theta^+\psi_+\psi_-$. Hence $N^2=2\theta^-\theta^+\psi_+\psi_-$ and $N^3=0$. The <exponential function> expands exactly as
$$
e^\Phi=e^\phi\left[1+i\theta^-\psi_++i\theta^+\psi_-+\theta^-\theta^+(iF+\psi_+\psi_-)\right].
$$
Equating the constant terms fixes the <auxiliary field>, $\boxed{F=e^\phi}$. The linear and quadratic terms then give the requested component field <equations of motion>:
$$
\boxed{\partial_-\psi_-=e^\phi\psi_+,\qquad \partial_+\psi_+=-e^\phi\psi_-,\qquad \partial_-\partial_+\phi=-e^{2\phi}+ie^\phi\psi_+\psi_-.}
$$
The signs here follow from the printed operator, the printed component expansion, and <left Grassmann derivatives>; no convention change has been made in eliminating the <auxiliary field>.
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