Simplify (2 - (y5)/(4)y(1)/(4) - y^-(1)/(4))

The denominator is $y^{\frac{1}{4}} - y^{-\frac{1}{4}}$.
Rewrite $y^{-\frac{1}{4}}$ as $\frac{1}{y^{\frac{1}{4}}}$:

$$y^{\frac{1}{4}} - \frac{1}{y^{\frac{1}{4}}}$$

Find a common denominator, which is $y^{\frac{1}{4}}$:

$$= \frac{y^{\frac{1}{4}} \cdot y^{\frac{1}{4}}}{y^{\frac{1}{4}}} - \frac{1}{y^{\frac{1}{4}}}$$ $$= \frac{y^{\frac{1}{4} + \frac{1}{4}} - 1}{y^{\frac{1}{4}}}$$ $$= \frac{y^{\frac{2}{4}} - 1}{y^{\frac{1}{4}}}$$ $$= \frac{y^{\frac{1}{2}} - 1}{y^{\frac{1}{4}}}$$

The fraction is $\frac{y^{\frac{5}{4}}}{\text{denominator}}$:

$$\frac{y^{\frac{5}{4}}}{\frac{y^{\frac{1}{2}} - 1}{y^{\frac{1}{4}}}}$$

Multiply the numerator by the reciprocal of the denominator:

$$= y^{\frac{5}{4}} \times \frac{y^{\frac{1}{4}}}{y^{\frac{1}{2}} - 1}$$

Combine the $y$ terms in the numerator using the exponent rule $a^m \times a^n = a^{m+n}$:

$$= \frac{y^{\frac{5}{4} + \frac{1}{4}}}{y^{\frac{1}{2}} - 1}$$ $$= \frac{y^{\frac{6}{4}}}{y^{\frac{1}{2}} - 1}$$

Simplify the exponent in the numerator: $\frac{6}{4} = \frac{3}{2}$.

$$= \frac{y^{\frac{3}{2}}}{y^{\frac{1}{2}} - 1}$$

The original expression is $2 - \text{fraction}$:

$$2 - \frac{y^{\frac{3}{2}}}{y^{\frac{1}{2}} - 1}$$

Find a common denominator, which is $(y^{\frac{1}{2}} - 1)$:

$$= \frac{2(y^{\frac{1}{2}} - 1)}{y^{\frac{1}{2}} - 1} - \frac{y^{\frac{3}{2}}}{y^{\frac{1}{2}} - 1}$$ $$= \frac{2(y^{\frac{1}{2}} - 1) - y^{\frac{3}{2}}}{y^{\frac{1}{2}} - 1}$$

Distribute the 2 in the numerator:

$$= \frac{2y^{\frac{1}{2}} - 2 - y^{\frac{3}{2}}}{y^{\frac{1}{2}} - 1}$$

Rearrange the terms in the numerator:

$$= \frac{-y^{\frac{3}{2}} + 2y^{\frac{1}{2}} - 2}{y^{\frac{1}{2}} - 1}$$

To make the leading term in the numerator positive, multiply the numerator and denominator by $-1$:

$$= \frac{-(y^{\frac{3}{2}} - 2y^{\frac{1}{2}} + 2)}{-(1 - y^{\frac{1}{2}})}$$ $$= \frac{y^{\frac{3}{2}} - 2y^{\frac{1}{2}} + 2}{1 - y^{\frac{1}{2}}}$$

The simplified expression is $\boxed{\frac{y^{\frac{3}{2}} - 2y^{\frac{1}{2}} + 2}{1 - y^{\frac{1}{2}}}}$.