There are two products that follow the same price $S(t)$ for all time $t$. The payout for product one is given by $w_1(\frac{100}{S(t)} - \frac{100}{S(t + \Delta t)})$ and the payout for product two is $w_2(S(t + \Delta t) - S(t))$. Where $w_1$ and $w_2$ are the quantities to buy or sell of product one and two respectively.
If at time 0 we buy $w_1$ units of product one at price $S(0)$, what quantity of $w_2$ of product two would we need to buy or sell also at time 0 such that our position is hedged? This means that for a small change in price, $\Delta S$, the change in total payout is zero. Your answer for $w_2$ should be a function of only $w_1$ and $S(0)$.
I have that the total payout is $$w_1\bigg(\frac{100}{S(t)} - \frac{100}{S(t + \Delta t)}\bigg) + w_2(S(t +\Delta t) - S(t))$$ Do I just have to take the derivative of this and set it to zero and solve for $w_2$? It does not seem right as I got $w_2 = \frac{-w_1 100}{S(0)^2}$.
Your idea works. For a small price change we can apply the Taylor approximation around $S(t)$:
$$\frac{100}{S(t+\Delta t)}\approx\frac{100}{S(t)}-\frac{100}{S(t)^2}(S(t+\Delta t)-S(t))$$
Plugging in to the payoff
$$\frac{100w_1}{S(t)^2}(S(t+\Delta t)-S(t))+w_2(S(t+\Delta t)-S(t))$$
$$=(S(t+\Delta t)-S(t))(\frac{100w_1}{S(t)^2}+w_2)$$
Hence if at time $t=0$ you set
$$w_2=-\frac{100w_1}{S(0)^2}$$ the position is hedged in that the payoff does not depend on small price change during $\Delta t$. This becomes exact in the continuous time limit and generally you need to adjust the position over time.