$\left(3xy\right)^0$

$\sin^2x=-1$

$\frac{d}{dx}\frac{x^2}{x+y}=y^2+9$

$\frac{-\left(-4+5-6\right)-\left(7-3+6\right)-5}{4\left(-1+3\right)-3}$

$\left(4x-1\right)\left(x+2\right)-4x^2+2$

$\lim_{x\to0}\left(\frac{\ln\left(x+1\right)+0.33\cdot\sin\left(3x\right)+\sin\left(2x\right)}{x^2}\right)$