The limit for problems like these do not exist since the limit depends on the direction you approach. For the problem you have mentioned, say you approach $(0,0,0)$ along the direction $y = m_y x$ and $z = m_z x$, where $m_y$, $m_z$ are some constants, then we get $\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \lim_{x \rightarrow 0} \frac{m_y x^2+2m_y m_z x^2 + 3m_z x^2}{x^2+4m_y^2 x^2+9 m_z^2 x^2}$ Hence, $\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \lim_{x \rightarrow 0} \frac{m_y + 2m_y m_z + 3m_z}{1 + 4 m_y^2 + 9 m_z^2} = \frac{m_y + 2m_y m_z + 3m_z}{1 + 4 m_y^2 + 9 m_z^2}$ You can key in different values for $m_y$ and $m_z$ and you will get different values.
Below are some instances of the different limits you get by approaching in different directions.
If you tend to zero along the direction $(1,0,0)$ i.e. $m_y = m_z = 0$, then we get $\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = 0.$
If you tend to zero along the direction $(1,1,0)$ i.e. $m_y = 1$ and $m_z = 0$, then we get $\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \frac15.$
If you tend to zero along the direction $(1,0,1)$ i.e. $m_y = 0$ and $m_z = 1$, then we get $\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \frac3{10}.$
If you tend to zero along the direction $(1,1,1)$ i.e. $m_y = 1$ and $m_z = 1$, then we get $\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \frac37.$