#### Answer

$-20y^{2}+3yx$

#### Work Step by Step

We are asked to subtract $(y^{2}+4yx+\frac{1}{7})$ from $(-19y^{2}+7yx+\frac{1}{7})$. This is equivalent to $(-19y^{2}+7yx+\frac{1}{7})-(y^{2}+4yx+\frac{1}{7})$.
To subtract two polynomials, we can add the opposite of the second polynomial to the first polynomial.
$(-19y^{2}+7yx+\frac{1}{7})+(-y^{2}-4yx-\frac{1}{7})$
Next, we can combine like terms.
$(-19y^{2}-y^{2})+(7yx-4yx)+(\frac{1}{7}-\frac{1}{7})=-20y^{2}+3yx$