Intersection of $36x^2 -9y^2+4z^2+36 = 0$ with plane $x=1$, derivative at a point The exercise asks me to find the inclination of the line tangent to the intersection of $36x^2 -9y^2+4z^2+36 = 0$ with the plane $x=1$ in the point $(1,\sqrt{12},3)$, and then say to me that I have to interpret this as a partial derivative.
So, that's what I did:
if we solve for $z$ in the equation, we have:
$$z = \pm \frac{\sqrt{9y^2-36x^2-36}}{2} \implies $$
if I take the partial with respect to $x$ for the $+$ function (because the point in $z$ is positive), all I'm doing is taking the partial derivative with respect to $x$, of the function defined by the intersection of that graph and the plaxe $x=1$, so I have:
$$\frac{\partial z}{\partial x} = \frac{-6x}{\sqrt{-4-4 x^2+y^2}}$$
that, in the point $(1,\sqrt(12),3)$ gives:
$$-3$$
Is everything ok?

UPDATE: I think now that this partial derivative should be with
  respect to $y$, rigth?

 A: Notice
Vector normal to the surface $f(x, y, z)=36x^2-9y^2+4z^2+36$ is given as $$\overrightarrow n=\left(i\frac{\partial f}{\partial x }+j\frac{\partial f}{\partial y }+k\frac{\partial f}{\partial z }\right)$$
$$=i(72x)+j(-18y)+k(8z)$$ Hence, the normal vector at the point of intersection $(1, \sqrt{12}, 3)$ is given as follows $$\overrightarrow n=i(72\times 1)+j(-18\sqrt {12})+k(8\times3 )$$$$=72i-36j\sqrt 3+24k$$ Now, the normal vector to the given plane $x=1$ is $i$ Hence, the angle say $\theta$ between the normal vector  to the surface & the normal vector to the plane: $$\theta=\cos^{-1}\left(\frac{(72i-36j\sqrt 3+24k)\cdot (i)}{\sqrt{72^2+(-36\sqrt 3)^2+24^2}\times 1}\right)$$
$$=\cos^{-1}\left(\frac{72}{6\sqrt{187}}\right)=\cos^{-1}\left(\frac{12}{\sqrt{187}}\right)\approx 28.65^\circ$$
Since, the normal & tangent are perpendicular to each other hence, the angle between the tangent to the curve & the plane $x=1$ is $$=90^\circ-\theta$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{90^\circ-\cos^{-1}\left(\frac{12}{\sqrt{187}}\right)\approx 61.34^\circ}}$$
A: For $x=1$, points on the curve satisfy $9y^2-4z^2=72$. Taking  this to define $z$ as a function of $y$ in a small neighborhood of $( \sqrt 12 ,3)$ ,we obtain $dz/dy = 3 \sqrt 3 /2$ when  $y=  \sqrt 12$,  which is therefore the partial  deriv of $z$ with respect to $y$ at the point $(1,\sqrt 12 ,3)$.The tangent line meets the plane $z=0$ at an angle $\arctan (3 \sqrt 3 /2)$.
A: For $x=1$, points on the curve satisfy $9y^2-4z^2=72$. Taking  this to define $z$ as a function of $y$ in a small neighborhood of $( \sqrt 12 ,3)$ ,we obtain $dz/dy = 3 \sqrt 3 /2$ when  $y=  \sqrt 12$,  which is the partial  deriv of $z$ with respect to $y$ at the point $(1,\sqrt 12 ,3)$.The tangent line meets the plane $z=0$ at an angle $\arctan (3 \sqrt 3 /2)$.
