Show that $\frac{\mathrm{d^{2}}B }{\mathrm{d} A^{2}}> 0 $ if $U''<0$. Given, $A = W_0 - L_0 + I - qI$, $B = W_0 - qI$, and $EU = p U(A) + (1-p) U(B) = k$, where $k$ is a constant. 
$\frac{\mathrm{d} B}{\mathrm{d} A}\bigg|_{}^{EU=k} = \frac{\frac{\partial EU }{\partial A}}{\frac{\partial EU}{\partial B}} = \frac{- p U'(A)}{(1-p) U'(B)}.$
The second derivative seems really complicated,as $A$ and $B$ are not separable. 
I think I may have to do implicit differentiation, so I would appreciate some help on that.
Edit: $$\frac{\mathrm{d^{2}}B }{\mathrm{d} A^{2}} = -\frac{\frac{\partial^2 EU }{\partial A^2}}{\frac{\partial^2 EU }{\partial B^2}} = -\frac{- p U''(A)}{(1-p) U''(B)} > 0.$$
I seem to be getting the right answer, but is my second derivative correct?
 A: As this appears to be an economics problem, I think an intuition approach is appropriate: suppose that $A$ goes up then $U'(A)$ goes down because $U'$ is decreasing. Moreover, $A\uparrow$ means $I\uparrow$ so $B$ decreases. This means $U'(B)$ goes up. Putting these together, you see that
$$
\frac{p U'(A)}{(1-p) U'(B)}
$$
goes down. Putting back the negative sign
$$
\frac{\mathrm{d} B}{\mathrm{d} A}\bigg|_{}^{EU=k} = \frac{- p U'(A)}{(1-p) U'(B)}
$$
goes up. In other words, $\frac{dB}{dA}$ is increasing in $A$, which is what you want.

Edit: A more rigorous approach: differentiate w.r.t. $A$
$$
k=p U(A)+(1-p)U(B)\implies 0=p U'(A)+(1-p)U'(B)\frac{dB}{dA}.
$$
Differentiate w.r.t. $A$ again
$$
0=p U''(A)+(1-p)U'(B)\frac{d^2B}{dA^2}+(1-p)U''(B)\left(\frac{dB}{dA}\right)^2
$$
from which you have
$$
\frac{d^2B}{dA^2}=-\frac{1}{(1-p)U'(B)}\left[p U''(A)+(1-p)U''(B)\left(\frac{dB}{dA}\right)^2\right]>0.
$$
If you want to get an explicit expression for $\frac{d^2B}{dA^2}$, you can substitute in $\frac{dB}{dA}=-\frac{p U'(A)}{(1-p) U'(B)}$.
