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. . .