Verified answer

[tex]\displaystyle\bf\\y=mx^{2} -x+m\\\\y=mx+1-m\\\\mx^{2} -x+m=mx+1-m\\\\mx^{2} -x-mx+m-1+m=0\\\\mx^{2} -(m+1)x+2m-1=0\\\\D=(m+1)^{2} -4\cdot m\cdot (2m-1)=m^{2} +2m+1-8m^{2} +4m=\\\\=-7m^{2} +6m+1\\\\D < 0\\\\-7m^{2} +6m+1 < 0\\\\7m^{2} -6m-1 > 0\\\\7\Big(m+\frac{1}{7} \Big)\Big(m-1\Big) > 0[/tex]

+ + + + + (- 1\7) - - - - - (1) + + + + +

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[tex]\displaystyle\bf\\m\in\Big(-\infty \ ; \ -\frac{1}{7} \Big) \ \cup \ \Big(1 \ ; \ +\infty\Big)[/tex]