LaGrange Multipliers Help 2

Here is a question I have been given in my exams:

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I have an answer but as I have no resources to check it against, I am posting my answers on here to try and get some extra help. Here is my answer:

a) $f(x,y,z)= \frac{x}{a}+\frac{y}{b}+ \frac{z}{c}, (\frac{x}{a})^2+(\frac{y}{b})^2+(\frac{z}{c})^2=1$

So $g(x,y,z)=(\frac{x}{a})^2+(\frac{y}{b})^2+(\frac{z}{c})^2$

$F_x={\lambda}g_x, \Longrightarrow \frac{1}{a}=\lambda(\frac{2x}{a^2}), \lambda=\frac{a}{2x}$ Call this equation (1).

$F_y={\lambda}g_y, \Longrightarrow \frac{1}{b}=\lambda(\frac{2y}{b^2}), \lambda=\frac{b}{2y}$ Call this equation (2).

(1)=(2) $\Longrightarrow \frac{a}{2x}=\frac{b}{2y} \Longrightarrow x=\frac{ay}{b}$

$F_z={\lambda}g_z, \Longrightarrow \frac{1}{c}=\lambda(\frac{2z}{c^2}), \lambda=\frac{c}{2z}$ Call this (3).

(3)=(2) $\Longrightarrow z=\frac{cy}{b}$

So:$(\frac{a\frac{y}{b}}{a})^2+(y/b)^2+(\frac{c\frac{y}{b}}{c})^2=1$

This gives:$\frac{3y^2}{b^2}=1, b= \pm \sqrt{3}y \Longrightarrow x=\frac{\pm a}{\sqrt{3}}, z=\frac{\pm c}{\sqrt{3}}$

I think this is the correct method, I do not however know how to tackle b. Thanks for any help!