Since the given equation
$\begin{equation*} x-\frac{x}{\sqrt{16-x^{2}}}-\frac{x}{\sqrt{9-x^{2}}}=0 \end{equation*}\tag{A}$
is equivalent to $\begin{equation*} x\underset{R}{\underbrace{\left( 1-\frac{1}{\sqrt{16-x^{2}}}-\frac{1}{\sqrt{9-x^{2}}}\right) }}=0,\tag{B} \end{equation*}$
$x=0$ is a root. The other roots are the roots of $R=0$. To illustrate the more general case we write it as $\begin{equation*} 1-\frac{1}{\sqrt{P}}-\frac{1}{\sqrt{Q}}=0, \end{equation*}\tag{C}$ where $\begin{equation*} P=16-x^{2},\qquad Q=9-x^{2}. \end{equation*}\tag{D}$
The usual method to solve (C) is
isolate one of the irrational terms on one side,
square both sides,$^1$
repeat until you have only rational terms,
transform the rational equation into a polynomial one.
For instance, starting with $-\frac{1}{\sqrt{P}}$ on the LHS and squaring, we get successively
$\begin{eqnarray*}-\frac{1}{\sqrt{P}}&=&\frac{1}{\sqrt{Q}}-1\Rightarrow \frac{1}{P} =\frac{1}{Q}-\frac{2}{\sqrt{Q}}+1,\qquad\text{(steps 1 and 2)}\\&\Leftrightarrow &\frac{2}{ \sqrt{Q}}=\frac{1}{Q}-\frac{1}{P}+1, \qquad\text{(steps 3 and 1) }\\ &\Rightarrow &\frac{4}{Q}=\frac{\left( P-Q+PQ\right) ^{2}}{P^{2}Q^{2}}, \qquad\text{(step 2)}\\&\Leftrightarrow &\frac{\left( P-Q+PQ\right) ^{2}}{P^{2}Q}=4,\qquad Q\neq 0, \\ &\Leftrightarrow &\left( P-Q+PQ\right) ^{2}-4P^{2}Q=0, \qquad\qquad\text{(step 4)}\qquad P,Q\neq 0,\\ &\Leftrightarrow &P^{2}Q^{2}+P^{2}-2PQ+Q^{2}-2P^{2}Q-2PQ^{2}=0,\qquad P,Q\neq 0.\tag{E} \end{eqnarray*}$
The degree of the last equation is defined by the 8th degree term: $\begin{equation*} P^{2}Q^{2}=\left( 16-x^{2}\right) ^{2}\left( 9-x^{2}\right) ^{2}\end{equation*}\tag{F}$
Substituting $P,Q$ and expanding yields the 8th. degree equation $\begin{equation*} x^{8}-46x^{6}+763x^{4}-5374x^{2}+13\,585=0.\tag{G} \end{equation*}$
However this particular equation can be reduced to a quartic (4th. degree) equation in $y=x^{2}$, which we could have expected because $R$ is a function of $x^2$
$\begin{equation*} y^{4}-46y^{3}+763y^{2}-5374y+13\,585=0.\tag{H} \end{equation*}$
This equation has two real roots $\begin{equation*} y_{1}\approx 6.7771,\qquad y_{2}\approx 8.4627. \end{equation*}$ The corresponding $x$ values are thus
$x_{1,2}\approx \pm \sqrt{6.7771}\approx 2.603\,3,\qquad x_{3,4}\approx \pm \sqrt{8.4627}\approx \pm 2.9091.$
Among these, $x_{3,4}$ are not solutions of the original equation (A) because
$\begin{equation*} 1-\frac{1}{\sqrt{16-y_{2}^2}}-\frac{1}{\sqrt{9-y_{2}^2}}\ne 0. \end{equation*}$
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$^1$ Raising both sides of an equation $A=B$ to the $n^{\text{th}}$ power yields a new equation $A^n=B^n$, which has all the solutions of the given equation and may admit other solutions, the so called extraneous solutions.
Case $n=2$. Since $A^{2}-B^{2}=(A-B)(A+B),$ the equation $A^{2}-B^{2}=0\Leftrightarrow A^{2}=B^{2}$ means that $A=B$ or $A=-B.$ The roots of the equation $A^2=B^2$ are the roots of the equation $A=B$ and the roots of the equation $A=-B$. In general the roots of equation $A=-B$ are different from the roots of the equation $A=B$, so when we square $A=B$ we may generate other solutions.