$y^2-2 x z\=1$

$y^2-2xz\=1$

$x^2-\cos \left(x z\right)\=\left(y\right)$

$x^2-\cos \left(xz\right)\=y$

$\mathrm{implicitdiff}\left(\left\{\,\right\}\, \left\{y\,z\right\}\, y\,x\right)$

$-\frac{2x}{\sin \left(xz\right)y-1}$

$\mathrm{implicitdiff}\left(\left\{\,\right\}\,\left\{y\,z\right\}\,z\,x\right)$

$-\frac{2yx\+z\sin \left(xz\right)y-z}{\left(\sin \left(xz\right)y-1\right)x}$

$\mathrm{eval}\left(\left[\,\right]\, \left[y\=y\left(x\right)\,z\=z\left(x\right)\right]\right)$

$\left[{y\left(x\right)}^2-2xz\left(x\right)\=1\,x^2-\cos \left(xz\left(x\right)\right)\=y\left(x\right)\right]$

$\mathrm{diff}\left(\,x\right)$

$\left[2y\left(x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-2z\left(x\right)-2x\left(\frac{\ⅆ}{\ⅆx}z\left(x\right)\right)\=0\,2x\+\sin \left(xz\left(x\right)\right)\left(z\left(x\right)\+x\left(\frac{\ⅆ}{\ⅆx}z\left(x\right)\right)\right)\=\frac{\ⅆ}{\ⅆx}y\left(x\right)\right]$

$\mathrm{solve}\left(\, \left[\mathrm{diff}\left(y\left(x\right)\,x\right)\,\mathrm{diff}\left(z\left(x\right)\,x\right)\right]\right)\left[1\right]$

$\left[\frac{\ⅆ}{\ⅆx}y\left(x\right)\=-\frac{2x}{\sin \left(xz\left(x\right)\right)y\left(x\right)-1}\,\frac{\ⅆ}{\ⅆx}z\left(x\right)\=-\frac{2y\left(x\right)x\+z\left(x\right)\sin \left(xz\left(x\right)\right)y\left(x\right)-z\left(x\right)}{\left(\sin \left(xz\left(x\right)\right)y\left(x\right)-1\right)x}\right]$

$⟨\frac{\mathrm{d\_\_y\_\_}}{\mathrm{d\_\_x\_\_}}\, \frac{\mathrm{d\_\_z\_\_}}{\mathrm{d\_\_x\_\_}}⟩\=\mathrm{eval}\left(⟨\mathrm{rhs}\left(\left[1\right]\right)\,\mathrm{rhs}\left(\left[2\right]\right)⟩\,\left[y\left(x\right)\=y\, z\left(x\right)\=z\right]\right)$

$\left[\begin{array}{c}\frac{\mathrm{dy}}{\mathrm{dx}}\\ \frac{\mathrm{dz}}{\mathrm{dx}}\end{array}\right]\=\left[\begin{array}{c}-\frac{2x}{\sin \left(xz\right)y-1}\\ -\frac{2yx\+z\sin \left(xz\right)y-z}{\left(\sin \left(xz\right)y-1\right)x}\end{array}\right]$