Consider two three-variable functions $H(x, y, z)$ and $K(x, y, z)$ and the associated level surfaces $H(x, y, z)= a$ and $K(x, y, z)= b.$ It is assumed that these surfaces intersect along some curve which contains the point $\langle x_0, y_0, z_0 \rangle.$ It is seen that this, presumable, defines a function $y= y(x)$ (e.g., the projection of this curve onto the xy-plane.) Derive the formula for $y'$ in the vicinity of $x_0$ in terms of the various partial derivatives of H and K- where it may be tacitly that appropriate combinations of these (which will appear in denominators) do not vanish.
Stared at this problem for awhile, but honestly have no idea how to even begin approaching this problem. Any suggestions/guidance toward beginning this problem would be very helpful.
Since you assume that these surfaces intersect to define a curve in space, then that curve can be parametrized by $x$ (at least in the vicinity of $x_0$), or in other words you have that $y = y(x)$ and $z = z(x)$ along this curve. This means that along the curve, $H$ and $K$ will be functions only of $x$, and moreover since they are constant everywhere, their derivatives with respect to $x$ must vanish:
$0 = \frac{d}{dx} H(x, y(x), z(x)) =\frac{\partial H}{\partial x}\Big|_{y,z} + \frac{\partial H}{\partial y}\Big|_{x,z}y'(x) + \frac{\partial H}{\partial z}\Big|_{x,y} z'(x)$
where the last equality above used the chain rule for partial derivatives. Doing the same thing for $K$ gives
$0 = \frac{d}{dx} K(x, y(x), z(x)) =\frac{\partial K}{\partial x}\Big|_{y,z} + \frac{\partial K}{\partial y}\Big|_{x,z}y'(x) + \frac{\partial K}{\partial z}\Big|_{x,y} z'(x)$
You are interested in evaluating all of these terms at $x_0$ of course. You now have two linear equations, both equal to zero, in the two unknowns $y'(x_0)$ and $z'(x_0)$ whose coefficients are the various partial derivatives indicated (again, evaluated at $(x_0, y(x_0), z(x_0)) = (x_0, y_0, z_0 $)), and you can solve for either one of them, in your case $y'(x_0)$, just by algebra.