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$\frac{1}{b(abc+a+c)} - \frac{1}{a+ \frac{1}{b+ \frac{1}{c} } } : \frac{1}{a+ \frac{1}{b} } = \frac{1}{b(abc+a+c)}- \frac{1}{a+ \frac{1}{ \frac{bc+1}{c} } } : \frac{1}{ \frac{ab+a}{b} } =\\ \frac{1}{b(abc+a+c)}- \frac{1}{a+1* \frac{c}{bc+1} } : 1* \frac{b}{ab+a} = \frac{1}{b(abc+a+c)}- \frac{1}{ \frac{a(bc+1)+c}{bc+1} } * \frac{ab+a}{b} =\\ \frac{1}{b(abc+a+c)}- \frac{bc+1}{a(bc+1)+c} * \frac{a(b+1)}{b} = \frac{1}{b(abc+a+c)}- \frac{(bc+1)*a(b+1)}{(abc+a+c)b} =\\ \frac{1-(bc+1)*a(b+1)}{b(abc+a+c)} =$ $\frac{1-(bc+1)(ab+1)}{b(abc+a+c)} = \frac{-b^ca-bc-ab}{b(abc+a+c)} = \frac{-b(ab+c+a)}{b(abc+a+c)} =- \frac{(ab+a+c)}{a(abc+a+c)}$