L'Hopital's rule is about evaluating limits of functions as the input approaches a value, so I assume you are wanting to calculate
#\lim_{x\to0} y#.
L'Hopitals rule deals with indeterminate forms of the type
#"f"(x)=\lim_{x \to a}("p"(x))/("q"(x)#,
where #"p"(x)# and #"q"(x)# are both indeterminate #(\pm\infty, \pminfty), (0,0),# as #x\toa#. In these cases L'Hopital's rule states
#\lim_{x \to a}("p"(x))/("q"(x)) = \lim_{x \to a}("p"'(x))/("q"'(x))#.
In this case, #y# is not in the form we require to use L'Hopital's rule. So we take the logarithm of both sides in order to find a form suitable for L'Hopital's rule.
#ln(y)=tan(3x)*ln(sin(2x))#
#ln(y)=ln(sin(2x))/(1/(tan(3x)))#
#ln(y)=ln(sin(2x))/cot(3x)#
#ln(sin(2x)) \to-\infty# and #cot(3x) \to\infty# as #x\to0#. Then this is an indeterminate form which could be resolved by L'Hopital's rule. Applying L'Hopital's rule,
#lim_{x\to0}ln(sin(2x))/cot(3x)=lim_{x\to0}((2cos(2x))/(sin(2x)))/(-3(1/sin^2(3x))#
#lim_{x\to0}ln(sin(2x))/cot(3x)=lim_{x\to0}(-2sin^2(3x)cos(2x))/(3sin(2x))#
#-2sin^2(3x)cos(2x) \to 0# and #3sin(2x) \to 0# as #x\to0#. Then this is an indeterminate form which could be resolved by L'Hopital's rule. Applying L'Hopital's rule,
#lim_{x\to0}(-2sin^2(3x)cos(2x))/(3sin(2x)) = lim_{x\to0}(-12sin(3x)cos(3x)cos(2x)+4sin^2(3x)sin(2x))/(6cos(2x))#
The right hand side can now be easily computed as #0#.
Then,
#lim_{x\to0}ln(y)=0#.
Taking an inverse logarithm,
#lim_{x\to\0}y=1#.
