\(C_3H_6O\) does not form a silver mirror with Tollen's reagent but forms an oxime with hydroxylamine. It can give positive:

Iodoform test

The correct answer is option 1. Iodoform test.

The given compound with the molecular formula \(C_3H_6O\) is identified as a ketone based on the fact that it forms an oxime with hydroxylamine. Ketones, along with aldehydes, react with hydroxylamine to form oximes. Since it does not form a silver mirror with Tollen's reagent, it indicates that the compound is not an aldehyde but rather a ketone. This conclusion is drawn because Tollen's reagent reacts specifically with aldehydes to form a silver mirror, while ketones do not undergo this reaction.

The structure , \(C_3H_6O\), represents common ketone, i.e., acetone. Acetone is a simple ketone consisting of two methyl groups (\(CH_3\)) attached to a carbonyl group (\(C=O\)).

Now, let's discuss the iodoform test. The iodoform test is a chemical test used to detect the presence of methyl ketones (\(CH_3CO\)) or methyl carbinols (\(CH_3CHOH\)). Ketones with a methyl group adjacent to the carbonyl carbon (\(CH_3CO\)) can undergo oxidation under basic conditions to form a carboxylate ion and iodoform (\(CHI_3\)). The iodoform forms a yellow precipitate, indicating a positive test result.

Given that acetone (\(CH_3COCH_3\)) contains a methyl ketone group (\(CH_3CO\)), it is expected to give a positive iodoform test. Therefore, the presence of a methyl ketone group in the structure of the compound supports the expectation of a positive iodoform test.

In summary, based on the structure (\(CH_3COCH_3\)) and the reactivity of the compound, it is reasonable to conclude that it can give a positive iodoform test. Hence, the correct answer is indeed the iodoform test, option (1).