Let there be the following substitutions:
$$\begin{align}
&w=x^2 &\\
&u=a+bw &j=d+fw
\end{align}$$
The equation reduces to:
$$\begin{align}
h&=\frac{u}{cx}\frac{j}{gx}\left(\frac{u}{cx}\frac{j}{gx}+i\right)-\left(\frac{u}{cx}\right)^2-\left(\frac{j}{gx}\right)^2\\
&=\left(\frac{uj}{cgx^2}\right)^2+\frac{uj}{cgx^2}i-\left(\frac{u}{cx}\right)^2-\left(\frac{j}{gx}\right)^2
\end{align}$$
Because of my slight dislike of fractional powers (I did not sub in $w^{\frac{1}{2}}$ for $x$), we can recall what $w$ is and restate the equation:
$$h=\frac{u^2j^2}{c^2g^2w^2}+\frac{uj}{cgw}i-\frac{u^2}{c^2w}-\frac{j^2}{g^2w}$$
Getting a polynomial in $w$ seems to be the most enlightening exercise, so multiply through by $w^2$:
$$hw^2=\frac{u^2j^2}{c^2g^2}+\frac{uj}{cg}iw-\frac{u^2}{c^2}w-\frac{j^2}{g^2}w$$
Rearranging this gives us a cleaner expression:
$$-hw^2-\left(\frac{u^2}{c^2}+\frac{j^2}{g^2}\right)w+\frac{uj}{cg}iw+\frac{u^2j^2}{c^2g^2}=0$$
Things will be slightly simpler if we multiply through by $c^2g^2$:
$$-hc^2g^2w^2-(u^2g^2+j^2c^2)w+ujcgwi+u^2j^2=0$$
Since $uj$, $u^2$, $j^2$ and $u^2j^2$ are the important things in the equation, let's simplify them:
$$\begin{align}
uj&=(a+bw)(d+fw)\\
&=bfw^2+(af+bd)w+ad\\
u^2&=(a+bw)^2\\
&=b^2w^2+2abw+a^2\\
j^2&=(d+fw)^2\\
&=f^2w^2+2dfw+d^2\\
u^2j^2&=(b^2w^2+2abw+a^2)(f^2w^2+2dfw+d^2)\\
&=(b^2f^2)w^4+(2abf^2+2b^2df)w^3+(a^2f^2+b^2d^2+4abdf)w^2+(2a^2df+2abd^2)w+(a^2d^2)
\end{align}
$$
It may help to work $u^2j^2$ with the following substitutions in mind:
$$\begin{align}
&\alpha_{2}=b^2 &\alpha_{1}=2ab &\alpha_{0}=a^2\\
&\beta_{2}=f^2 &\beta_{1}=2df &\alpha_{0}=d^2
\end{align}$$
The equivalent equation is, thusly:
$$u^2j^2=\alpha_2 \beta_2 w^4+(\alpha_1 \beta_{2}+\alpha_{2} \beta_{1})w^3+(\alpha_0 \beta_2+\alpha_1 \beta_1+\alpha_2 \beta_0)w^2+(\alpha_0 \beta_1+\alpha_1 \beta_0)w+(\alpha_0 \beta_0)$$
It is now a task of digging through this algebra to separate the terms. . .
Firstly, look at the 2nd coefficient:
$$\begin{align}
u^2g^2+j^2c^2&=(b^2w^2+2abw+a^2)g^2+(f^2w^2+2dfw+d^2)c^2\\
&=(b^2g^2+f^2c^2)w^2+(2abg^2+2dfc^2)w+(a^2g^2+d^2c^2)
\end{align}$$
Next, look at the third term:
$$\begin{align}
ujcgwi&=[bfw^2+(af+bd)w+ad]cgwi\\
&=(bfcgi)w^3+(afcgi+bdcgi)w^2+(adcgi)w
\end{align}$$
Now, combining all of these forms is the nearly final step (Let the equation be abbreviated with an E):
$$
\begin{align}
E&=-hc^2g^2w^2-((b^2g^2+f^2c^2)w^2+(2abg^2+2dfc^2)w+(a^2g^2+d^2c^2))w\\
&+(bfcgi)w^3+(afcgi+bdcgi)w^2+(adcgi)w\\
&+(b^2f^2)w^4+(2abf^2+2b^2df)w^3\\
&+(a^2f^2+b^2d^2+4abdf)w^2+(2a^2df+2abd^2)w+(a^2d^2)\\
&=(b^2f^2)w^4\\
&+(2abf^2+2b^2df)w^3-(b^2g^2+f^2c^2)w^3++(bfcgi)w^3\\
&+(a^2f^2+b^2d^2+4abdf)w^2-(2abg^2+2dfc^2)w^2+(afcgi+bdcgi)w^2-(hc^2g^2)w^2\\
&-(a^2g^2+d^2c^2)w+(2a^2df+2abd^2)w+(adcgi)w\\
&+a^2d^2
\end{align}
$$
This shows us that the coefficients of the terms are as follows:
$$
\begin{align}
C_4&=b^2f^2\\
C_3&=2abf^2+2b^2df-b^2g^2-f^2c^2+bfcgi\\
C_2&=a^2f^2+b^2d^2+4abdf-2abg^2-2dfc^2+afcgi+bdcgi-hc^2g^2\\
C_1&=-a^2g^2-d^2c^2+2a^2df+2abd^2+adcgi\\
C_0&=a^2d^2
\end{align}
$$
Thus, we have the following quartic in $w$:
$$C_4w^4+C_3w^3+C_2w^2+C_1w+C_0=0$$
I leave the rest as an exercise to the reader. . .