1-sin6x-cos6x/1-sin4x-cos4x
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(1-sin6x-cos6x)/(1-sin4x-cos4x)

=[1-(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)]/(1-sin^4x-cos^4x)

=(1-sin^4x-cos^4x+sin^2xcos^2x)/(1-sin^4x-cos^4x)

=(1-1+sin^2xcos^2x+sin^2xcos^2x)/(1-1+2sin^2xcos^2x)

=3sin^2xcos^2x/2sin^2xcos^2x

=3/2