Let's remember that a tangent line is another phrase for the word "slope" or derivative in the lingo of calculus. Since the tangent line is expected to be horizontal, then the slope is 0 (from the days of algebra/algebra 2).
But, since we know that slope is another way to talk about a derivative, then let's take the derivative and set it to zero...
y = x2ln(x)
dy/dx = y' = x2 * 1/x + 2x*ln(x) ---> by way of the product rule
dy/dx = x + 2x*ln(x) = 0
Factor out x to see what we have...
x*(1 + 2ln(x)) = 0
Since we have two factors, either of them could be zero --> x = 0, or 1 + 2ln(x) = 0. Attempting to solve the second equation (i.e. root) gives...
2ln(x) = -1 --> ln(x) = -1/2 --> x = e-1/2 = 1/√e
So we have two answers --> x = 0, or x = 1/√e which will yield a horizontal tangent line
The second problem uses the concept of position, velocity, and acceleration...
velocity = derivative of position with respect to time --> v = df/dt
acceleration = derivative of the velocity with respect to time --> a = dv/dt = d2f/dt2
f = 3cos(2t)
f' = v = 3 * 2 * (-sin(2t)) = -6sin(2t)
f" = a = -6 * 2 * (cos(2t)) = -12cos(2t) = 4 * f