by gradient i think you mean slope ie #dy/dx#. [there is a big difference though it comes later in the math journey, when gradient does not mean slope!!]
We can say, from the chain rule, that
#dy/dx = dy/dt*dt/dx = (dy/dt)/(dx/dt) = (-2 sin 2t)/(- 3 sin t)#
#= (4 sin t cos t)/( 3 sin t)# ....we just used the ID, #sin 2A = 2 sin A cos A#
#= 4/3 cos t#
Now if
#[dy/dx]_{t = t_o} = 1 #
then
# 4/3 cos t_o = 1 implies cos t_o = 3/4#
from the Pythagorean ID: #sin^2 + cos ^2 = 1#
we know that #sin t_o = sqrt 7 /4#
the point on the curve is # ((x), (y)) = ((3 cos t), (cos 2t))#
and from an additional trig ID: #cos 2A = 2cos^2 A - 1#
we can say that #cos 2 t_o = 2 * 9/16 -1 = 2/16#
so
# ((x), (y)) = ((3 * 3/4), (1/8)) = ((9/4), (1/8)) #
this can be also done by de-parameterising the equation
from #x = 3 cos t#, we say that #cos t = x/3#
we then have #y = cos 2t# and from that double angle identity we say that
#y = 2 cos^2 t - 1#
#= 2 (x/3)^2 - 1#
so #y' = (4x)/3*1/3 = (4x)/9#
#y' = 1 implies x = 9/4#
and so
#y = 2 ((9/4)/3)^2 - 1 = 18/16 -1 = 1/8#
