= Solution
Write the surfaces as level sets $f_1=x^2y^2z-3x-z^2=0$ and $f_2=3x^2-y^2+2z-1=0$. Their <gradients> are <normal vectors> to their tangent planes. At the specified point they are $\mathbf n_1=(5,-4,2)$ and $\mathbf n_2=(6,4,2)$, with dot product $18$ and squared norms $45,56$. The <angle between intersecting regular surfaces> is the angle between their tangent planes, so
$$
\boxed{\theta=\arccos\frac{|18|}{\sqrt{45}\sqrt{56}}=\arccos\frac3{\sqrt{70}}\simeq68.99^\circ.}
$$
Both gradients are nonzero, ensuring that the tangent planes are defined.
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