Find Max Value for each month for the last 3 months, properly
For SQL Server, you could do something like this.

DECLARE @t TABLE (Id INT, DateVal DATETIME, ValueVal DECIMAL(18, 9));

INSERT @t ( Id, DateVal, ValueVal )
SELECT Id, DateVal, ValueVal
FROM (
    VALUES 
    (12410,   '01/03/2017 12:17',    0.000178),
    (12410,   '01/10/2017 11:36',    0.000186),
    (12410,   '01/17/2017 11:27',    0.000189),
    (12410,   '01/24/2017 13:09',    0.000182),
    (12410,   '01/31/2017 10:37',    0.000169),
    (12410,   '02/07/2017 11:03',    0.000214),
    (12410,   '02/14/2017 11:52',    0.000176),
    (12410,   '02/21/2017 10:51',    0.000200),
    (12410,   '02/28/2017 12:29',    0.000194),
    (12410,   '03/07/2017 08:39',    0.000206)
) x (Id, DateVal, ValueVal);

SELECT  DATEPART(MONTH, t.DateVal) AS [DateVal],
        MAX(t.ValueVal) AS MaxVal
FROM @t AS t
WHERE t.DateVal &gt;= DATEADD(MONTH, -3, DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0))
GROUP BY DATEPART(MONTH, t.DateVal);


Updated WHERE clause:
Flattening dates is weird, and my previous query got you today's date minus three months. To get back to the first of three months ago, you have to turn some tricks.

SELECT DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0) AS [First Of This Month],
       DATEADD(MONTH, -3, DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0)) AS [First Of Three Months Ago]


You can either take some time trying to understand this, or keep a cheat sheet of how to do it handy so you don't have to remember awful date math ;)