I am looking at a problem where a biased coin with probability for heads $p_h=0.6$ is flipped a large number of times (e.g. $n = 100000$). Now I am asked to find the boundaries within which the relative number of heads will lie with a probability of 90%. Am I right to assume that I have to derive the confidence interval for a bernoulli-trial and calculate the 90% CI around 0.60, which in my case would be $\text{CI} = [0.60-\frac{1.645}{100000}{\sqrt{0.60(1-0.60)}},0.60+\frac{1.645}{100000}{\sqrt{0.60(1-0.60)}}]$?
As far as the derivation of the CI for the Bernoulli case is concerned, I have already found helpful material, I just wanted to make sure I correctly understand the exercise in this context.
It indeed sounds like the problem is asking you to find a 90% confidence interval for the sample proportion $\hat{\pi}$. Your confidence interval is almost correct. The sample sizes $n=100000$ should be under the square root because the standard deviation of a sample proportion $\hat{\pi}$ (from an underlying binomial distribution) is asymptotically $\sigma(\hat{\pi})=\sqrt{\frac{\pi(1-\pi)}{n}}$.
Thus the C.I. should be
$$\left(0.6-1.645\sqrt{\frac{0.6(1-0.6)}{100000}},0.6+1.645\sqrt{\frac{0.6(1-0.6)}{100000}}\right)=(0.5975,0.6025).$$