Let #(x_1,y_1)=(pi/2,pi/4)#
By implicitly differentiating the left-hand side with respect to #x#,
#d/{dx}(y sin12x)#
by Product Rule,
#={dy}/{dx}cdot sin12x+y cdot 12cos12x#
#=sin12x{dy}/{dx}+12ycos12x#
by plugging in #(x,y)=(pi/2,pi/4)#,
#=sin(6pi){dy}/{dx}+3picos(6pi)=3pi#
By implicitly differentiating the right-hand side with respect to #x#,
#d/{dx}(x cos2y)#
by Product Rule,
#=1cdot cos2y+x cdot(-sin2y)(2{dy}/{dx})#
#=cos2y-2xsin2y{dy}/{dx}#
by plugging in #(x,y)=(pi/2,pi/4)#,
#=cos(pi/2)-pi sin(pi/2){dy}/{dx}=-pi{dy}/{dx}#
Now, by setting the left-hand side and the right-hand side equal to each other,
#-pi{dy}/{dx}=3pi#
So, the slope #m# of the tangent line is:
#m={dy}/{dx}|_{(pi/2,pi/4)}=-3#
By Point-Slope Form #y-y_1=m(x-x_1)#,
#y-pi/4=-3(x-pi/2)#
I hope that this was helpful.
