<s>[INST] Is $\frac{x^2-3x-4}{-3x-15}\Big/\frac{x^2-16}{x^2-x-30}$ defined at $x=6$ or not? If a number is in the denominator of a rational expression that itself is the denominator of another rational expression, such that the number causes the latter denominator to be undefined, then isn't the entire expression undefined?
For example, isn't $$\frac{\frac{x^2-3x-4}{-3x-15}}{\frac{x^2-16}{x^2-x-30}}=\frac{\frac{(x-4)(x+1)}{-3(x+5)}}{\frac{(x+4)(x-4)}{(x-6)(x+5)}}$$ undefined at $x=6\;?$ $$\frac{\frac{x^2-3x-4}{-3x-15}}{\frac{x^2-16}{x^2-x-30}} \implies\quad\frac{(x-4)(x+1)(x-6)(x+5)}{-3(x+5)(x+4)(x-4)},x\neq{-5},{-4},4,6\quad?$$
If so, why do my scientific calculator and Desmos compute the expression as zero at $x=6\;?$
 [/INST] 
$$\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}} \implies\quad\frac{(x-4)(x+1)(x-6)(x+5)}{-3(x+5)(x+4)(x-4)},x\neq{-5},{-4},4,6\quad?$$

Rather: $$f(x)=\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}}\iff\bigg(\color\red{x\neq-5,4,6}\quad\text{and}\quad f(x)=\frac{(x+1)(x-6)}{-3(x+4)} \bigg).$$

isn't $$\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}}$$ undefined at $x=6\;?$

This expression is indeed undefined at $x=6;$ Wolfram also agrees.

why do my scientific calculator and Desmos compute the expression as zero at $x=6\;?$

Most likely, Desmos determines a graphing function's domain only after getting rid of any compound (multi-storey) fraction: $$\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}}$$ contains four implicit conditions $(\color\red{x\ne{-}5,{-}4,4,6}),$ while $$\frac{(x-4)(x+1)(x-6)(x+5)}{-3(x+5)(x+4)(x-4)}$$ contains only three implicit conditions $(\color\red{x\ne{-}5,{-}4,4}).$
Because of this, Desmos is selectively noticing the compound fraction's implicit conditions, mistaking $f(6)$ as $0$ even while it doesn't similarly think that $f(-5)= \frac{44}3$ or that $f(4)=\frac5{12}.$
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